2.1 The Tangent and Velocity Problems
1. (a) Using (15 250), we construct the following table:
slope =
5 (5 694) 694 5−−15 250 = − 444 10 = −444
10 (10 444) 444 10−−250 15 = − 194 5 = −388
20 (20 111
...
2.1 The Tangent and Velocity Problems
1. (a) Using (15 250), we construct the following table:
slope =
5 (5 694) 694 5−−15 250 = − 444 10 = −444
10 (10 444) 444 10−−250 15 = − 194 5 = −388
20 (20 111) 111 20−−250 15 = − 139 5 = −278
25 (25 28) 28 25−−250 15 = − 222 10 = −222
30 (30 0) 0 30 −−250 15 = − 250 15 = −166
(b) Using the values of that correspond to the points
closest to ( = 10 and = 20), we have
−388 + (−278)
2
= −333
(c) From the graph, we can estimate the slope of the
tangent line at to be −300 9 = −333.
2. (a) Slope = 2948 42 −− 2530 36 = 418 6 ≈ 6967 (b) Slope = 2948 42 − − 2661 38 = 287 4 = 7175
(c) Slope = 2948 42 −− 2806 40 = 142 2 = 71 (d) Slope = 3080 44 − − 2948 42 = 132 2 = 66
From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeatsminute after 42 minutes.
After being stable for a while, the patient’s heart rate is dropping.
3. (a) = 1
1 − , (2 −1)
( 1(1 − ))
(i) 15 (15 −2) 2
(ii) 19 (19 −1111 111) 1111 111
(iii) 199 (199 −1010 101) 1010 101
(iv) 1999 (1999 −1001 001) 1001 001
(v) 25 (25 −0666 667) 0666 667
(vi) 21 (21 −0909 091) 0909 091
(vii) 201 (201 −0990 099) 0990 099
(viii) 2001 (2001 −0999 001) 0999 001
(b) The slope appears to be 1.
(c) Using = 1, an equation of the tangent line to the
curve at (2 −1) is − (−1) = 1( − 2), or
= − 3.
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4. (a) = cos , (05 0)
(i) 0 (0 1) −2
(ii) 04 (04 0309017) −3090170
(iii) 049 (049 0031411) −3141076
(iv) 0499 (0499 0003142) −3141587
(v) 1 (1 −1) −2
(vi) 06 (06 −0309017) −3090170
(vii) 051 (051 −0031411) −3141076
(viii) 0501 (0501 −0003142) −3141587
(b) The slope appears to be −.
(c) − 0 = −( − 05) or = − + 1 2.
(d)
5. (a) = () = 40 − 162. At = 2, = 40(2) − 16(2)2 = 16. The average velocity between times 2 and 2 + is
ave =
(2 + ) − (2)
(2 + ) − 2 =
40(2 + ) − 16(2 + )2 − 16
=
−24 − 162
= −24 − 16, if 6= 0.
(i) [2 25]: = 05, ave = −32 fts (ii) [2 21]: = 01, ave = −256 fts
(iii) [2 205]: = 005, ave = −248 fts (iv) [2 201]: = 001, ave = −2416 fts
(b) The instantaneous velocity when = 2 ( approaches 0) is −24 fts.
6. (a) = () = 10 − 1862. At = 1, = 10(1) − 186(1)2 = 814. The average velocity between times 1 and 1 + is
ave =
(1 + ) − (1)
(1 + ) − 1 =
10(1 + ) − 186(1 + )2 − 814
=
628 − 1862
= 628 − 186, if 6= 0.
(i) [1 2]: = 1, ave = 442 ms (ii) [1 15]: = 05, ave = 535 ms
(iii) [1 11]: = 01, ave = 6094 ms (iv) [1 101]: = 001, ave = 62614 ms
(v) [1 1001]: = 0001, ave = 627814 ms
(b) The instantaneous velocity when = 1 ( approaches 0) is 628 ms.
7. (a) (i) On the interval [2 4] , ave = (4) − (2)
4 − 2 =
792 − 206
2
= 293 fts.
(ii) On the interval [3 4] , ave = (4) − (3)
4 − 3 =
792 − 465
1
= 327 fts.
(iii) On the interval [4 5] , ave = (5) − (4)
5 − 4 =
1248 − 792
1
= 456 fts.
(iv) On the interval [4 6] , ave = (6) − (4)
6 − 4 =
1767 − 792
2
= 4875 fts.
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(b) Using the points (2 16) and (5 105) from the approximate
tangent line, the instantaneous velocity at = 3 is about
105 − 16
5 − 2
=
89
3
≈ 297 fts.
8. (a) (i) = () = 2 sin + 3 cos . On the interval [1 2], ave = (2) − (1)
2 − 1
=
3 − (−3)
1
= 6 cms.
(ii) On the interval [1 11], ave = (11) − (1)
11 − 1
≈
−3471 − (−3)
01
= −471 cms.
(iii) On the interval [1 101], ave = (101) − (1)
101 − 1
≈
−30613 − (−3)
001
= −613 cms.
(iv) On the interval [1 1001], ave = (1001) − (1)
1001 − 1
≈
−300627 − (−3)
0001
= −627 cms.
(b) The instantaneous velocity of the particle when = 1 appears to be about −63 cms.
9. (a) For the curve = sin(10) and the point (1 0):
2 (2 0) 0
15 (15 08660) 17321
14 (14 −04339) −10847
13 (13 −08230) −27433
12 (12 08660) 43301
11 (11 −02817) −28173
05 (05 0) 0
06 (06 08660) −21651
07 (07 07818) −26061
08 (08 1) −5
09 (09 −03420) 34202
As approaches 1, the slopes do not appear to be approaching any particular value.
(b) We see that problems with estimation are caused by the frequent
oscillations of the graph. The tangent is so steep at that we need to
take -values much closer to 1 in order to get accurate estimates of
its slope.
(c) If we choose = 1001, then the point is (1001 −00314) and ≈ −313794. If = 0999, then is
(0999 00314) and = −314422. The average of these slopes is −314108. So we estimate that the slope of the
tangent line at is about −314.
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2.2 The Limit of a Function
1. As approaches 2, () approaches 5. [Or, the values of () can be made as close to 5 as we like by taking sufficiently
close to 2 (but 6= 2).] Yes, the graph could have a hole at (2 5) and be defined such that (2) = 3.
2. As approaches 1 from the left, () approaches 3; and as approaches 1 from the right, () approaches 7. No, the limit
does not exist because the left- and right-hand limits are different.
3. (a) lim
→−3
() = ∞ means that the values of () can be made arbitrarily large (as large as we please) by taking
sufficiently close to −3 (but not equal to −3).
(b) lim
→4+
() = −∞ means that the values of () can be made arbitrarily large negative by taking sufficiently close to 4
through values larger than 4.
4. (a) As approaches 2 from the left, the values of () approach 3, so lim
→2−
() = 3.
(b) As approaches 2 from the right, the values of () approach 1, so lim
→2+
() = 1.
(c) lim
→2
() does not exist since the left-hand limit does not equal the right-hand limit.
(d) When = 2, = 3, so (2) = 3.
(e) As approaches 4, the values of () approach 4, so lim
→4
() = 4.
(f ) There is no value of () when = 4, so (4) does not exist.
5. (a) As approaches 1, the values of () approach 2, so lim
→1
() = 2.
(b) As approaches 3 from the left, the values of () approach 1, so lim
→3−
() = 1.
(c) As approaches 3 from the right, the values of () approach 4, so lim
→3+
() = 4.
(d) lim
→3
() does not exist since the left-hand limit does not equal the right-hand limit.
(e) When = 3, = 3, so (3) = 3.
6. (a) () approaches 4 as approaches −3 from the left, so lim
→−3−
() = 4.
(b) () approaches 4 as approaches −3 from the right, so lim
→−3+
() = 4.
(c) lim
→−3
() = 4 because the limits in part (a) and part (b) are equal.
(d) (−3) is not defined, so it doesn’t exist.
(e) () approaches 1 as approaches 0 from the left, so lim
→0−
() = 1.
(f ) () approaches −1 as approaches 0 from the right, so lim
→0+
() = −1.
(g) lim
→0
() does not exist because the limits in part (e) and part (f ) are not equal.
(h) (0) = 1 since the point (0 1) is on the graph of .
(i) Since lim
→2−
() = 2 and lim
→2+
() = 2, we have lim
→2
() = 2.
(j) (2) is not defined, so it doesn’t exist.
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(k) () approaches 3 as approaches 5 from the right, so lim
→5+
() = 3.
(l) () does not approach any one number as approaches 5 from the left, so lim
→5−
() does not exist.
7. (a) lim
→0−
() = −1 (b) lim
→0+
() = −2
(c) lim
→0
() does not exist because the limits in part (a) and part (b) are not equal.
(d) lim
→2−
() = 2 (e) lim
→2+
() = 0
(f) lim
→2
() does not exist because the limits in part (d) and part (e) are not equal.
(g) (2) = 1 (h) lim
→4
() = 3
8. (a) lim
→−3
() = ∞ (b) lim
→2
() does not exist. (c) lim
→2−
() = −∞
(d) lim
→2+
() = ∞ (e) lim
→−1
() = −∞
(f) The equations of the vertical asymptotes are = −3, = −1 and = 2.
9. (a) lim
→−7
() = −∞ (b) lim
→−3
() = ∞ (c) lim
→0
() = ∞
(d) lim
→6−
() = −∞ (e) lim
→6+
() = ∞
(f) The equations of the vertical asymptotes are = −7, = −3, = 0, and = 6.
10. lim
→12−
() = 150 mg and lim
→12+
() = 300 mg. These limits show that there is an abrupt change in the amount of drug in
the patient’s bloodstream at = 12 h. The left-hand limit represents the amount of the drug just before the fourth injection.
The right-hand limit represents the amount of the drug just after the fourth injection.
11. From the graph of
() =
1 + if −1
2 if −1 ≤ 1
2 − if ≥ 1
,
we see that lim
→
() exists for all except = −1. Notice that the
right and left limits are different at = −1.
12. From the graph of
() =
1 + sin if 0
cos if 0 ≤ ≤
sin if
,
we see that lim
→
() exists for all except = . Notice that the
right and left limits are different at = .
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13. (a) lim
→0−
() = 1
(b) lim
→0+
() = 0
(c) lim
→0
() does not exist because the limits
in part (a) and part (b) are not equal.
14. (a) lim
→0−
() = −1
(b) lim
→0+
() = 1
(c) lim
→0
() does not exist because the limits
in part (a) and part (b) are not equal.
15. lim
→0−
() = −1, lim
→0+
() = 2, (0) = 1 16. lim
→0
() = 1, lim
→3−
() = −2, lim
→3+
() = 2,
(0) = −1, (3) = 1
17. lim
→3+
() = 4, lim
→3−
() = 2, lim
→−2
() = 2,
(3) = 3, (−2) = 1
18. lim
→0−
() = 2, lim
→0+
() = 0, lim
→4−
() = 3,
lim
→4+
() = 0, (0) = 2, (4) = 1
19. For () = 2 − 3
2 − 9 :
()
31 0508 197
305 0504 132
301 0500 832
3001 0500 083
30001 0500 008
()
29 0491 525
295 0495 798
299 0499 165
2999 0499 917
29999 0499 992
It appears that lim
→3
2 − 3
2 − 9 =
1 2
.
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20. For () = 2 − 3
2 − 9 :
()
−25 −5
−29 −29
−295 −59
−299 −299
−2999 −2999
−29999 −29,999
()
−35 7
−31 31
−305 61
−301 301
−3001 3001
−30001 30,001
It appears that lim
→−3+
() = −∞ and that
lim
→−3−
() = ∞, so lim
→−3
2 − 3
2 − 9 does not exist.
21. For () = 5 − 1
:
()
05 22364 988
01 6487 213
001 5127 110
0001 5012 521
00001 5001 250
()
−05 1835 830
−01 3934 693
−001 4877 058
−0001 4987 521
−00001 4998 750
It appears that lim
→0
5 − 1
= 5.
22. For () = (2 + )5 − 32
:
()
05 131312 500
01 88410 100
001 80804 010
0001 80080 040
00001 80008 000
()
−05 48812 500
−01 72390 100
−001 79203 990
−0001 79920 040
−00001 79992 000
It appears that lim
→0
(2 + )5 − 32
= 80.
23. For () = ln − ln 4
− 4 :
()
39 0253 178
399 0250 313
3999 0250 031
39999 0250 003
()
41 0246 926
401 0249 688
4001 0249 969
40001 0249 997
It appears that lim
→4
() = 025. The graph confirms that result.
24. For () = 1 + 9
1 + 15 :
()
−11 0427 397
−101 0582 008
−1001 0598 200
−10001 0599 820
()
−09 0771 405
−099 0617 992
−0999 0601 800
−09999 0600 180
It appears that lim
→−1
() = 06. The graph confirms that result.
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25. For () = sin 3
tan 2 :
()
±01 1457 847
±001 1499 575
±0001 1499 996
±00001 1500 000
It appears that lim
→0
sin 3
tan 2 = 15.
The graph confirms that result.
26. For () = 5 − 1
:
()
01 1746 189
001 1622 459
0001 1610 734
00001 1609 567
()
−01 1486 601
−001 1596 556
−0001 1608 143
−00001 1609 308
It appears that lim
→0
() ≈ 16094. The graph confirms that result.
27. For () = :
()
01 0794 328
001 0954 993
0001 0993 116
00001 0999 079
It appears that lim
→0+
() = 1.
The graph confirms that result.
28. For () = 2 ln :
()
01 −0023 026
001 −0000 461
0001 −0000 007
00001 −0000 000
It appears that lim
→0+
() = 0.
The graph confirms that result.
29. (a) From the graphs, it seems that lim
→0
cos 2 − cos
2 = −15. (b)
()
±01 −1493 759
±001 −1499 938
±0001 −1499 999
±00001 −1500 000
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30. (a) From the graphs, it seems that lim
→0
sin
sin
= 032. (b)
()
±01 0323 068
±001 0318 357
±0001 0318 310
±00001 0318 310
Later we will be able to show that
the exact value is 1
.
31. lim
→5+
+ 1
− 5
= ∞ since the numerator is positive and the denominator approaches 0 from the positive side as → 5+.
32. lim
→5−
+ 1
− 5
= −∞ since the numerator is positive and the denominator approaches 0 from the negative side as → 5−.
33. lim
→1
2 −
( − 1)2 = ∞ since the numerator is positive and the denominator approaches 0 through positive values as → 1.
34. lim
→3−
√
( − 3)5 = −∞ since the numerator is positive and the denominator approaches 0 from the negative side as → 3−.
35. Let = 2 − 9. Then as → 3+, → 0+, and lim
→3+
ln(2 − 9) = lim
→0+
ln = −∞ by (5).
36. lim
→0+
ln(sin ) = −∞ since sin → 0+ as → 0+.
37. lim
→(2)+
1
sec = −∞ since 1
is positive and sec → −∞ as → (2)+.
38. lim
→−
cot = lim
→−
cos
sin = −∞ since the numerator is negative and the denominator approaches 0 through positive values
as → −.
39. lim
→2−
csc = lim
→2−
sin
= −∞ since the numerator is positive and the denominator approaches 0 through negative
values as → 2−.
40. lim
→2−
2 − 2
2 − 4 + 4 = lim →2− ((−−2)2) 2 = lim →2− − 2 = −∞ since the numerator is positive and the denominator
approaches 0 through negative values as → 2−.
41. lim
→2+
2 − 2 − 8
2 − 5 + 6 = lim →2+ ( ( − − 4)( 3)( + 2) − 2) = ∞ since the numerator is negative and the denominator approaches 0 through
negative values as → 2+.
42. lim
→0+ 1 − ln = ∞ since 1 → ∞ and ln → −∞ as → 0+.
43. lim
→0
(ln 2 − −2) = −∞ since ln 2 → −∞ and −2 → ∞ as → 0.
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44. (a) The denominator of = 2 + 1
3 − 22 =
2 + 1
(3 − 2) is equal to zero when
= 0 and = 3
2 (and the numerator is not), so = 0 and = 15 are
vertical asymptotes of the function.
(b)
45. (a) () = 1
3 − 1.
From these calculations, it seems that
lim
→1−
() = −∞ and lim
→1+
() = ∞.
()
05 −114
09 −369
099 −337
0999 −3337
09999 −33337
099999 −33,3337
()
15 042
11 302
101 330
1001 3330
10001 33330
100001 33,3333
(b) If is slightly smaller than 1, then 3 − 1 will be a negative number close to 0, and the reciprocal of 3 − 1, that is, (),
will be a negative number with large absolute value. So lim
→1−
() = −∞.
If is slightly larger than 1, then 3 − 1 will be a small positive number, and its reciprocal, (), will be a large positive
number. So lim
→1+
() = ∞.
(c) It appears from the graph of that
lim
→1−
() = −∞ and lim
→1+
() = ∞.
46. (a) From the graphs, it seems that lim
→0
tan 4
= 4. (b)
()
±01 4227 932
±001 4002 135
±0001 4000 021
±00001 4000 000
47. (a) Let () = (1 + )1.
()
−0001 271964
−00001 271842
−000001 271830
−0000001 271828
0000001 271828
000001 271827
00001 271815
0001 271692
It appears that lim
→0
(1 + )1 ≈ 271828, which is approximately .
In Section 3.6 we will see that the value of the limit is exactly .
(b)
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48. (a)
No, because the calculator-produced graph of () = + ln | − 4| looks like an exponential function, but the graph of
has an infinite discontinuity at = 4. A second graph, obtained by increasing the numpoints option in Maple, begins to
reveal the discontinuity at = 4.
(b) There isn’t a single graph that shows all the features of . Several graphs are needed since looks like ln | − 4| for large
negative values of and like for 5, but yet has the infinite discontiuity at = 4.
A hand-drawn graph, though distorted, might be better at revealing the main
features of this function.
49. For () = 2 − (21000):
(a)
()
1 0998 000
08 0638 259
06 0358 484
04 0158 680
02 0038 851
01 0008 928
005 0001 465
It appears that lim
→0
() = 0.
(b)
()
004 0000 572
002 −0000 614
001 −0000 907
0005 −0000 978
0003 −0000 993
0001 −0001 000
It appears that lim
→0
() = −0001.
50. For () = tan −
3 :
(a)
()
10 0557 407 73
05 0370 419 92
01 0334 672 09
005 0333 667 00
001 0333 346 67
0005 0333 336 67
(b) It seems that lim
→0
() = 1 3.
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© Cengage Learning. All Rights Reserved.78 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
(c)
()
0001 0333 333 50
00005 0333 333 44
00001 0333 330 00
000005 0333 336 00
000001 0333 000 00
0000001 0000 000 00
Here the values will vary from one
calculator to another. Every calculator will
eventually give false values.
(d) As in part (c), when we take a small enough viewing rectangle we get incorrect output.
51. No matter how many times we zoom in toward the origin, the graphs of () = sin() appear to consist of almost-vertical
lines. This indicates more and more frequent oscillations as → 0.
52. (a) For any positive integer , if = 1
, then () = tan 1
= tan() = 0. (Remember that the tangent function has
period .)
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© Cengage Learning. All Rights Reserved.SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 79
(b) For any nonnegative number , if = 4
(4 + 1) , then
() = tan 1
= tan
(4 + 1)
4 = tan 4 4 + 4 = tan + 4 = tan 4 = 1
(c) From part (a), () = 0 infinitely often as → 0. From part (b), () = 1 infinitely often as → 0. Thus, lim
→0
tan
1
does not exist since () does not get close to a fixed number as → 0.
53. There appear to be vertical asymptotes of the curve = tan(2 sin ) at ≈ ±090
and ≈ ±224. To find the exact equations of these asymptotes, we note that the
graph of the tangent function has vertical asymptotes at = 2 + . Thus, we
must have 2 sin =
2 + , or equivalently, sin = 4 + 2 . Since
−1 ≤ sin ≤ 1, we must have sin = ± 4 and so = ± sin−1 4 (corresponding
to ≈ ±090). Just as 150◦ is the reference angle for 30◦, − sin−1 4 is the
reference angle for sin−1 4 . So = ± − sin−1 4 are also equations of
vertical asymptotes (corresponding to ≈ ±224).
54. lim
→−
= lim
→−
0
1 − 22 . As → −, 1 − 22 → 0+, and → ∞.
55. (a) Let = √3−−11.
From the table and the graph, we guess
that the limit of as approaches 1 is 6.
099 5925 31
0999 5992 50
09999 5999 25
101 6075 31
1001 6007 50
10001 6000 75
(b) We need to have 55 √3−−11 65. From the graph we obtain the approximate points of intersection (09314 55)
and (10649 65). Now 1 − 09314 = 00686 and 10649 − 1 = 00649, so by requiring that be within 00649 of 1,
we ensure that is within 05 of 6.
2.3 Calculating Limits Using the Limit Laws
1. (a) lim
→2
[() + 5()] = lim
→2
() + lim
→2
[5()] [Limit Law 1]
= lim
→2
() + 5 lim
→2
() [Limit Law 3]
= 4 + 5(−2) = −6
(b) lim
→2
[()]3 = lim →2 ()3 [Limit Law 6]
= ( −2)3 = −8
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(c) lim
→2
() = lim →2 () [Limit Law 11]
= √4 = 2
(d) lim
→2
3()
() =
lim
→2
[3()]
lim
→2
() [Limit Law 5]
=
3 lim
→2
()
lim
→2
() [Limit Law 3]
=
3(4)
−2
= −6
(e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim
→2
()
(), does not exist because the
denominator approaches 0 while the numerator approaches a nonzero number.
(f) lim
→2
() ()
() =
lim
→2
[() ()]
lim
→2
() [Limit Law 5]
=
lim
→2
() · lim
→2
()
lim
→2
() [Limit Law 4]
=
−2 · 0
4 = 0
2. (a) lim
→2
[() + ()] = lim
→2
() + lim
→2
() [Limit Law 1]
= −1 + 2
= 1
(b) lim
→0
() exists, but lim
→0
() does not exist, so we cannot apply Limit Law 2 to lim
→0
[() − ()].
The limit does not exist.
(c) lim
→−1
[() ()] = lim
→−1
() · lim
→−1
() [Limit Law 4]
= 1 · 2
= 2
(d) lim
→3
() = 1, but lim
→3
() = 0, so we cannot apply Limit Law 5 to lim
→3
()
() . The limit does not exist.
Note: lim
→3−
()
() = ∞ since () → 0+ as → 3− and lim →3+ (()) = −∞ since () → 0−as → 3+.
Therefore, the limit does not exist, even as an infinite limit.
(e) lim
→2
2() = lim
→2
2 · lim
→2
() [Limit Law 4]
= 22 · (−1)
= −4
(f) (−1) + lim
→−1
() is undefined since (−1) is
not defined.
3. lim
→3
(53 − 32 + − 6) = lim
→3
(53) − lim
→3
(32) + lim
→3
− lim
→3
6 [Limit Laws 2 and 1]
= 5 lim
→3
3 − 3 lim
→3
2 + lim
→3
− lim
→3
6 [3]
= 5(33) − 3(32) + 3 − 6 [9, 8, and 7]
= 105
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4. lim
→−1
(4 − 3)(2 + 5 + 3) = lim
→−1
(4 − 3) lim
→−1
(2 + 5 + 3) [Limit Law 4]
= lim →−1 4 − lim →−1 3lim →−1 2 + lim →−1 5 + lim →−1 3 [2, 1]
= lim →−1 4 − 3 lim →−1 lim →−1 2 + 5 lim →−1 + lim →−1 3 [3]
= (1 + 3)(1 − 5 + 3) [9, 8, and 7]
= 4(−1) = −4
5. lim
→−2
4 − 2
22 − 3 + 2 =
lim
→−2
(4 − 2)
lim
→−2
(22 − 3 + 2) [Limit Law 5]
=
lim
→−2
4 − lim
→−2
2
2 lim
→−2
2 − 3 lim
→−2
+ lim
→−2
2
[1, 2, and 3]
=
16 − 2
2(4) − 3(−2) + 2 [9, 7, and 8]
=
14
16
=
7 8
6. lim
→−2
√4 + 3 + 6 = lim →−2 (4 + 3 + 6) [11]
= lim →−2 4 + 3 lim →−2 + lim →−2 6 [1, 2, and 3]
= (−2)4 + 3 (−2) + 6 [9, 8, and 7]
= √16 − 6 + 6 = √16 = 4
7. lim
→8
(1 + √3 ) (2 − 62 + 3) = lim
→8
(1 + √3 ) · lim
→8
(2 − 62 + 3) [Limit Law 4]
= lim →8 1 + lim →8 √3 · lim →8 2 − 6 lim →8 2 + lim →8 3 [1, 2, and 3]
= 1 + √3 8 · 2 − 6 · 82 + 83 [7, 10, 9]
= (3)(130) = 390
8. lim
→23 −2 3− 2+ 52 = lim →2 3 −2 3− 2+ 52 [Limit Law 6]
=
lim
→2
(2 − 2)
lim
→2
(3 − 3 + 5)
2
[5]
=
lim
→2
2 − lim
→2
2
lim
→2
3 − 3 lim
→2
+ lim
→2
5
2
[1, 2, and 3]
= 8 −43(2) + 5 − 2 2 [9, 7, and 8]
= 722 = 49 4
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© Cengage Learning. All Rights Reserved.82 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
9. lim
→2 232−+ 1 2 = lim →2 232−+ 1 2 [Limit Law 11]
=
lim
→2
(22 + 1)
lim
→2
(3 − 2) [5]
=
2 lim
→2
2 + lim
→2
1
3 lim
→2
− lim
→2
2
[1, 2, and 3]
= 2(2) 3(2)2−+ 1 2 = 9 4 = 32 [9, 8, and 7]
10. (a) The left-hand side of the equation is not defined for = 2, but the right-hand side is.
(b) Since the equation holds for all 6= 2, it follows that both sides of the equation approach the same limit as → 2, just as
in Example 3. Remember that in finding lim
→
(), we never consider = .
11. lim
→5
2 − 6 + 5
− 5
= lim
→5
( − 5)( − 1)
− 5
= lim
→5
( − 1) = 5 − 1 = 4
12. lim
→−3
2 + 3
2 − − 12 = lim →−3 ( −(4)( + 3) + 3) = lim →−3 − 4 = −3−−3 4 = 37
13. lim
→5
2 − 5 + 6
− 5
does not exist since − 5 → 0, but 2 − 5 + 6 → 6 as → 5.
14. lim
→4
2 + 3
2 − − 12 = lim →4 ( −(4)( + 3) + 3) = lim →4 − 4. The last limit does not exist since lim →4− − 4 = −∞ and
lim
→4+
− 4 = ∞.
15. lim
→−3
2 − 9
22 + 7 + 3 = lim →−3 (2 (+ 3)( + 1)(−+ 3) 3) = lim →−3 2−+ 1 3 = 2(−−33) + 1 − 3 = −−65 = 65
16. lim
→−1
22 + 3 + 1
2 − 2 − 3 = lim →−1 (2 (−+ 1)( 3)(+ 1) + 1) = lim →−1 2−+ 1 3 = 2(−−11) + 1 − 3 = −−14 = 14
17. lim
→0
(−5 + )2 − 25
= lim →0 (25 − 10+ 2) − 25 = lim →0 −10+ 2 = lim →0 (−10 + ) = lim →0(−10 + ) = −10
18. lim
→0
(2 + )3 − 8
= lim →0
8 + 12 + 62 + 3 − 8
= lim →0 12 + 62 + 3
= lim
→0
12 + 6 + 2 = 12 + 0 + 0 = 12
19. By the formula for the sum of cubes, we have
lim
→−2
+ 2
3 + 8 = lim →−2 ( + 2)(2+ 2 − 2 + 4) = lim →−2 2 − 21 + 4 = 4 + 4 + 4 1 = 12 1 .
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© Cengage Learning. All Rights Reserved.SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 83
20. We use the difference of squares in the numerator and the difference of cubes in the denominator.
lim
→1
4 − 1
3 − 1 = lim →1 ((−2 1)( − 1)( 2 +2 + 1) + 1) = lim →1 ((− −1)( 1)( + 1)( 2 + 2+ 1) + 1) = lim →1 ( + 1)( 2 + 2+ 1 + 1) = 2(2) 3 = 43
21. lim
→0
√9 + − 3
= lim →0
√9 + − 3
·
√9 + + 3
√9 + + 3 = lim →0
√9 + 2 − 32
√9 + + 3 = lim →0 (9 + √9 +)−+ 3 9
= lim
→0
√9 + + 3 = lim →0 √9 +1 + 3 = 3 + 3 1 = 1 6
22. lim
→2
√4 + 1 − 3
− 2
= lim
→2
√4 + 1 − 3
− 2
·
√4 + 1 + 3
√4 + 1 + 3 = lim →2
√4 + 1 2 − 32
( − 2)√4 + 1 + 3
= lim
→2
4 + 1 − 9
( − 2)√4 + 1 + 3 = lim →2 ( − 2)4( √4−2) + 1 + 3
= lim
→2
4
√4 + 1 + 3 =
4
√9 + 3 =
2 3
23. lim
→3
1
−
1 3
− 3
= lim
→3
1
−
1 3
− 3
·
3
3
= lim
→3
3 −
3( − 3) = lim →3 −31 = −19
24. lim
→0
(3 + )−1 − 3−1
= lim →0
1
3 + −
1 3
= lim →0 3−(3 + (3 +)3 ) = lim →0 (3 + −)3
= lim
→0 −3(3 + 1 ) = − lim 1
→0
[3(3 + )] = −
1
3(3 + 0) = −
1 9
25. lim
→0
√1 + − √1 −
= lim
→0
√1 + − √1 −
·
√1 + + √1 −
√1 + + √1 − = lim →0
√1 + 2 − √1 − 2
√1 + + √1 −
= lim
→0
(1 + ) − (1 − )
√1 + + √1 − = lim →0 √1 + 2+ √1 − = lim →0 √1 + +2 √1 −
=
2
√1 + √1 =
2 2
= 1
26. lim
→0 1 − 2 1+ = lim →0 1 − ( + 1) 1 = lim →0 + 1 ( + 1) − 1 = lim →0 + 1 1 = 0 + 1 1 = 1
27. lim
→16
4 − √
16 − 2 = lim →16 (16 (4 − −√2)(4 + )(4 +√√)) = lim →16 (16 −16 )(4 + − √ )
= lim
→16
1
(4 + √ ) =
1
16 4 + √16 =
1
16(8) =
1
128
28. lim
→2
2 − 4 + 4
4 − 32 − 4 = lim →2 (2 −(4)( − 2)22+ 1) = lim →2 ( + 2)( ( − − 2) 2)(22 + 1)
= lim
→2
− 2
( + 2)(2 + 1) =
0
4 · 5 = 0
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© Cengage Learning. All Rights Reserved.84 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
29. lim
→0 √1 + 1 − 1 = lim →0 1−√√1 + 1 + = lim →0 1−√√ + 1 1 +1 + 1 + √1 + √1 + = lim →0 √1 + 1 + − √1 +
= lim
→0
−1
√1 + 1 + √1 + =
−1
√1 + 0 1 + √1 + 0 = −
1 2
30. lim
→−4
√2 + 9 − 5
+ 4 = lim →−4
√2 + 9 − 5√2 + 9 + 5
( + 4)√2 + 9 + 5 = lim →−4 ( + 4) (2+ 9) √2 −+ 9 + 5 25
= lim
→−4
2 − 16
( + 4)√2 + 9 + 5 = lim →−4 ( + 4) ( + 4)( √2+ 9 + 5 − 4)
= lim
→−4
− 4
√2 + 9 + 5 =
−4 − 4
√16 + 9 + 5 =
−8
5 + 5
= −
4 5
31. lim
→0
( + )3 − 3
= lim →0 (3 + 32 + 32 + 3) − 3 = lim →0 32 + 32 + 3
= lim
→0
(32 + 3 + 2)
= lim →0(32 + 3 + 2) = 32
32. lim
→0
1
( + )2 −
1 2
= lim →0
2 − ( + )2
( + )22
= lim →0 2 − (22(+ 2 + )2+ 2) = lim →0 −2(2 (++))2
= lim
→0
−(2 + )
2( + )2 =
−2
2 · 2 = −
2 3
33. (a)
lim
→0
√1 + 3 − 1 ≈
2 3
(b)
()
−0001 0666 166 3
−0000 1 0666 616 7
−0000 01 0666 661 7
−0000 001 0666 666 2
0000 001 0666 667 2
0000 01 0666 671 7
0000 1 0666 716 7
0001 0667 166 3
The limit appears to be 2
3
.
(c) lim
→0 √1 + 3 − 1 · √√1 + 3 1 + 3 + 1 + 1 = lim →0 (1 + 3 √1 + 3 )−+ 1 1 = lim →0 √1 + 3 3 + 1
=
1 3
lim
→0
√1 + 3 + 1 [Limit Law 3]
=
1 3
lim →0(1 + 3) + lim →0 1 [1 and 11]
=
1 3
lim →0 1 + 3 lim →0 + 1 [1, 3, and 7]
=
1 3
√1 + 3 · 0 + 1 [7 and 8]
=
1 3
(1 + 1) = 2
3
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© Cengage Learning. All Rights Reserved.SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 85
34. (a)
lim
→0
√3 + − √3
≈ 029
(b)
()
−0001 0288 699 2
−0000 1 0288 677 5
−0000 01 0288 675 4
−0000 001 0288 675 2
0000 001 0288 675 1
0000 01 0288 674 9
0000 1 0288 672 7
0001 0288 651 1
The limit appears to be approximately 02887.
(c) lim
→0 √3 + − √3 · √√3 + 3 + + + √ √3 3 = lim →0 √(3 + 3 +)+−√3 3 = lim →0 √3 + 1 + √3
=
lim
→0
1
lim
→0
√3 + + lim
→0
√3 [Limit Laws 5 and 1]
=
1
lim →0 (3 + ) + √3 [7 and 11]
=
1
√3 + 0 + √3 [1, 7, and 8]
=
1
2 √3
35. Let () = −2, () = 2 cos 20 and () = 2. Then
−1 ≤ cos 20 ≤ 1 ⇒ −2 ≤ 2 cos 20 ≤ 2 ⇒ () ≤ () ≤ ().
So since lim
→0
() = lim
→0
() = 0, by the Squeeze Theorem we have
lim
→0
() = 0.
36. Let () = −√3 + 2, () = √3 + 2 sin(), and () = √3 + 2. Then
−1 ≤ sin() ≤ 1 ⇒ −√3 + 2 ≤ √3 + 2 sin() ≤ √3 + 2 ⇒
() ≤ () ≤ (). So since lim
→0
() = lim
→0
() = 0, by the Squeeze Theorem
we have lim
→0
() = 0.
37. We have lim
→4
(4 − 9) = 4(4) − 9 = 7 and lim
→4
2 − 4 + 7 = 42 − 4(4) + 7 = 7. Since 4 − 9 ≤ () ≤ 2 − 4 + 7
for ≥ 0, lim
→4
() = 7 by the Squeeze Theorem.
38. We have lim
→1
(2) = 2(1) = 2 and lim
→1
(4 − 2 + 2) = 14 − 12 + 2 = 2. Since 2 ≤ () ≤ 4 − 2 + 2 for all ,
lim
→1
() = 2 by the Squeeze Theorem.
39. −1 ≤ cos(2) ≤ 1 ⇒ −4 ≤ 4 cos(2) ≤ 4. Since lim
→0
−4 = 0 and lim
→0
4 = 0, we have
lim
→0
4 cos(2) = 0 by the Squeeze Theorem.
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40. −1 ≤ sin() ≤ 1 ⇒ −1 ≤ sin() ≤ 1 ⇒ √/ ≤ √ sin() ≤ √ . Since lim
→0+
(√/) = 0 and
lim
→0+
(√ ) = 0, we have lim
→0+ √ sin() = 0 by the Squeeze Theorem.
41. | − 3| = − (−3− 3) if if − − 3 3 ≥ 0 0 = 3 −−3 if if ≥ 33
Thus, lim
→3+
(2 + | − 3|) = lim
→3+
(2 + − 3) = lim
→3+
(3 − 3) = 3(3) − 3 = 6 and
lim
→3−
(2 + | − 3|) = lim
→3−
(2 + 3 − ) = lim
→3−
( + 3) = 3 + 3 = 6. Since the left and right limits are equal,
lim
→3
(2 + | − 3|) = 6.
42. | + 6| = − (+ 6 + 6) if if + 6 + 6 ≥ 0 0 = −(+ 6 + 6) if if ≥ −−66
We’ll look at the one-sided limits.
lim
→−6+
2 + 12
| + 6| = lim →−6+ 2(+ 6 + 6) = 2 and →− lim6− 2| + 12 + 6| = lim →−6− −2((+ 6) + 6) = −2
The left and right limits are different, so lim
→−6
2 + 12
| + 6| does not exist.
43. 23 − 2 = 2(2 − 1) = 2 · |2 − 1| = 2 |2 − 1|
|2 − 1| = −2(2−1− 1) if if 2 2 − − 1 1 ≥ 0 0 = 2 −(2−1− 1) if if ≥ 0055
So 23 − 2 = 2[−(2 − 1)] for 05.
Thus, lim
→05−
2 − 1
|23 − 2| = lim →05− 2[−2(2− −1 1)] = lim →05− −21 = (0−5) 1 2 = 0−25 1 = −4.
44. Since || = − for 0, we have lim
→−2
2 − ||
2 +
= lim
→−2
2 − (−)
2 +
= lim
→−2
2 +
2 +
= lim
→−2
1 = 1.
45. Since || = − for 0, we have lim
→0− 1 − |1| = lim →0− 1 − −1 = lim →0− 2 , which does not exist since the
denominator approaches 0 and the numerator does not.
46. Since || = for 0, we have lim
→0+ 1 − |1| = lim →0+ 1 − 1 = lim →0+ 0 = 0.
47. (a) (b) (i) Since sgn = 1 for 0, lim
→0+
sgn = lim
→0+
1 = 1.
(ii) Since sgn = −1 for 0, lim
→0−
sgn = lim
→0−
−1 = −1.
(iii) Since lim
→0−
sgn 6= lim
→0+
sgn , lim
→0
sgn does not exist.
(iv) Since |sgn | = 1 for 6= 0, lim
→0
|sgn | = lim
→0
1 = 1.
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© Cengage Learning. All Rights Reserved.SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 87
48. (a) () = sgn(sin ) =
−1 if sin 0
0 if sin = 0
1 if sin 0
(i) lim
→0+
() = lim
→0+
sgn(sin ) = 1 since sin is positive for small positive values of .
(ii) lim
→0−
() = lim
→0−
sgn(sin ) = −1 since sin is negative for small negative values of .
(iii) lim
→0
() does not exist since lim
→0+
() 6= lim
→0−
().
(iv) lim
→+
() = lim
→+
sgn(sin ) = −1 since sin is negative for values of slightly greater than .
(v) lim
→−
() = lim
→−
sgn(sin ) = 1 since sin is positive for values of slightly less than .
(vi) lim
→
() does not exist since lim
→+
() 6= lim
→−
().
(b) The sine function changes sign at every integer multiple of , so the
signum function equals 1 on one side and −1 on the other side of ,
an integer. Thus, lim
→
() does not exist for = , an integer.
(c)
49. (a) (i) lim
→2+
() = lim
→2+
2 + − 6
| − 2| = lim →2+ ( + 3)( | −2|− 2)
= lim
→2+
( + 3)( − 2)
− 2
[since − 2 0 if → 2+]
= lim
→2+
( + 3) = 5
(ii) The solution is similar to the solution in part (i), but now | − 2| = 2 − since − 2 0 if → 2−.
Thus, lim
→2−
() = lim
→2−
−( + 3) = −5.
(b) Since the right-hand and left-hand limits of at = 2
are not equal, lim
→2
() does not exist.
(c)
50. (a) () =(2 −+ 1 2)2 if if ≥ 1 1
lim
→1−
() = lim
→1−
(2 + 1) = 12 + 1 = 2, lim
→1+
() = lim
→1+
( − 2)2 = (−1)2 = 1
(b) Since the right-hand and left-hand limits of at = 1
are not equal, lim
→1
() does not exist.
(c)
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51. For the lim
→2
() to exist, the one-sided limits at = 2 must be equal. lim
→2−
() = lim
→2−
4 − 1 2 = 4 − 1 = 3 and
lim
→2+
() = lim
→2+
√ + = √2 + . Now 3 = √2 + ⇒ 9 = 2 + ⇔ = 7.
52. (a) (i) lim
→1−
() = lim
→1−
= 1
(ii) lim
→1+
() = lim
→1+
(2 − 2) = 2 − 12 = 1. Since lim
→1−
() = 1 and lim
→1+
() = 1, we have lim
→1
() = 1.
Note that the fact (1) = 3 does not affect the value of the limit.
(iii) When = 1, () = 3, so (1) = 3.
(iv) lim
→2−
() = lim
→2−
(2 − 2) = 2 − 22 = 2 − 4 = −2
(v) lim
→2+
() = lim
→2+
( − 3) = 2 − 3 = −1
(vi) lim
→2
() does not exist since lim
→2−
() 6= lim
→2+
().
(b)
() =
if 1
3 if = 1
2 − 2 if 1 ≤ 2
− 3 if 2
53. (a) (i) [[]] = −2 for −2 ≤ −1, so lim
→−2+
[[]] = lim
→−2+
(−2) = −2
(ii) [[]] = −3 for −3 ≤ −2, so lim
→−2−
[[]] = lim
→−2−
(−3) = −3.
The right and left limits are different, so lim
→−2
[[]] does not exist.
(iii) [[]] = −3 for −3 ≤ −2, so lim
→−24
[[]] = lim
→−24
(−3) = −3.
(b) (i) [[]] = − 1 for − 1 ≤ , so lim
→−
[[]] = lim
→−
( − 1) = − 1.
(ii) [[]] = for ≤ + 1, so lim
→+
[[]] = lim
→+
= .
(c) lim
→
[[]] exists ⇔ is not an integer.
54. (a) See the graph of = cos .
Since −1 ≤ cos 0 on [− −2), we have = () = [[cos ]] = −1
on [− −2).
Since 0 ≤ cos 1 on [−2 0) ∪ (0 2], we have () = 0
on [−2 0) ∪ (0 2].
Since −1 ≤ cos 0 on (2 ], we have () = −1 on (2 ].
Note that (0) = 1.
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© Cengage Learning. All Rights Reserved.SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 89
(b) (i) lim
→0−
() = 0 and lim
→0+
() = 0, so lim
→0
() = 0.
(ii) As → (2)−, () → 0, so lim
→(2)−
() = 0.
(iii) As → (2)+, () → −1, so lim
→(2)+
() = −1.
(iv) Since the answers in parts (ii) and (iii) are not equal, lim
→2
() does not exist.
(c) lim
→
() exists for all in the open interval (− ) except = −2 and = 2.
55. The graph of () = [[]] + [[−]] is the same as the graph of () = −1 with holes at each integer, since () = 0 for any
integer . Thus, lim
→2−
() = −1 and lim
→2+
() = −1, so lim
→2
() = −1. However,
(2) = [[2]] + [[−2]] = 2 + (−2) = 0, so lim
→2
() 6= (2).
56. lim
→− 01 − 22 = 0√1 − 1 = 0. As the velocity approaches the speed of light, the length approaches 0.
A left-hand limit is necessary since is not defined for .
57. Since () is a polynomial, () = 0 + 1 + 22 + · · · + . Thus, by the Limit Laws,
lim
→
() = lim
→
0 + 1 + 22 + · · · + = 0 + 1 lim
→
+ 2 lim
→
2 + · · · + lim
→
= 0 + 1 + 22 + · · · + = ()
Thus, for any polynomial , lim
→
() = ().
58. Let () = ()
() where () and () are any polynomials, and suppose that () 6= 0. Then
lim
→
() = lim
→
()
() =
lim
→
()
lim
→
() [Limit Law 5] = (()) [Exercise 57] = ().
59. lim
→1
[() − 8] = lim
→1 (−) −1 8 · ( − 1) = lim →1 (−) −1 8 · lim →1( − 1) = 10 · 0 = 0.
Thus, lim
→1
() = lim
→1
{[() − 8] + 8} = lim
→1
[() − 8] + lim
→1
8 = 0 + 8 = 8.
Note: The value of lim
→1
() − 8
− 1
does not affect the answer since it’s multiplied by 0. What’s important is that
lim
→1
() − 8
− 1
exists.
60. (a) lim
→0
() = lim
→0 (2) · 2 = lim →0 (2) · lim →0 2 = 5 · 0 = 0
(b) lim
→0
()
= lim
→0 (2) · = lim →0 (2) · lim →0 = 5 · 0 = 0
61. Observe that 0 ≤ () ≤ 2 for all , and lim
→0
0 = 0 = lim
→0
2. So, by the Squeeze Theorem, lim
→0
() = 0.
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62. Let () = [[]] and () = −[[]]. Then lim
→3
() and lim
→3
() do not exist [Example 10]
but lim
→3
[() + ()] = lim
→3
([[]] − [[]]) = lim
→3
0 = 0.
63. Let () = () and () = 1 − (), where is the Heaviside function defined in Exercise 1.3.59.
Thus, either or is 0 for any value of . Then lim
→0
() and lim
→0
() do not exist, but lim
→0
[()()] = lim
→0
0 = 0.
64. lim
→2
√6 − − 2
√3 − − 1 = lim →2 √ √6 3 − − − − 2 1 · √√6 6 − − + 2 + 2 · √√3 3 − − + 1 + 1
= lim
→2 √√6 3 − − 2 2 − − 2 12 2 · √√3 6 − − + 1 + 2 = lim →2 6 3 − − − − 4 1 · √√3 6 − − + 1 + 2
= lim
→2
(2 − )√3 − + 1
(2 − )√6 − + 2 = lim →2
√3 − + 1
√6 − + 2 =
1 2
65. Since the denominator approaches 0 as → −2, the limit will exist only if the numerator also approaches
0 as → −2. In order for this to happen, we need lim
→−2
32 + + + 3 = 0 ⇔
3(−2)2 + (−2) + + 3 = 0 ⇔ 12 − 2 + + 3 = 0 ⇔ = 15. With = 15, the limit becomes
lim
→−2
32 + 15 + 18
2 + − 2 = lim →−2 3( (−+ 2)( 1)(+ 2) + 3) = lim →−2 3(−+ 3) 1 = 3(−−22 + 3) − 1 = −33 = −1.
66. Solution 1: First, we find the coordinates of and as functions of . Then we can find the equation of the line determined
by these two points, and thus find the -intercept (the point ), and take the limit as → 0. The coordinates of are (0 ).
The point is the point of intersection of the two circles 2 + 2 = 2 and ( − 1)2 + 2 = 1. Eliminating from these
equations, we get 2 − 2 = 1 − ( − 1)2 ⇔ 2 = 1 + 2 − 1 ⇔ = 1 22. Substituting back into the equation of the
shrinking circle to find the -coordinate, we get 1 222 + 2 = 2 ⇔ 2 = 21 − 1 42 ⇔ = 1 − 1 42
(the positive -value). So the coordinates of are 1 22 1 − 1 42 . The equation of the line joining and is thus
− =
1 − 1 42 −
12
2 − 0
( − 0). We set = 0 in order to find the -intercept, and get
= −
12
2
1 − 1 42 − 1 =
−
12
2 1 − 1 42 + 1
1 − 1
42 − 1
= 2 1 − 1 42 + 1
Now we take the limit as → 0+: lim
→0+
= lim
→0+
21 − 1 42 + 1= lim →0+ 2 √1 + 1 = 4.
So the limiting position of is the point (4 0).
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Solution 2: We add a few lines to the diagram, as shown. Note that
∠ = 90◦ (subtended by diameter ). So ∠ = 90◦ = ∠
(subtended by diameter ). It follows that ∠ = ∠ . Also
∠ = 90◦ − ∠ = ∠ . Since 4 is isosceles, so is
4, implying that = . As the circle 2 shrinks, the point
plainly approaches the origin, so the point must approach a point twice
as far from the origin as , that is, the point (4 0), as above.
2.4 The Precise Definition of a Limit
1. If |() − 1| 02, then −02 () − 1 02 ⇒ 08 () 12. From the graph, we see that the last inequality is
true if 07 11, so we can choose = min {1 − 07 11 − 1} = min {03 01} = 01 (or any smaller positive
number).
2. If |() − 2| 05, then −05 () − 2 05 ⇒ 15 () 25. From the graph, we see that the last inequality is
true if 26 38, so we can take = min {3 − 26 38 − 3} = min {04 08} = 04 (or any smaller positive number).
Note that 6= 3.
3. The leftmost question mark is the solution of √ = 16 and the rightmost, √ = 24. So the values are 162 = 256 and
242 = 576. On the left side, we need | − 4| |256 − 4| = 144. On the right side, we need | − 4| |576 − 4| = 176.
To satisfy both conditions, we need the more restrictive condition to hold— namely, | − 4| 144. Thus, we can choose
= 144, or any smaller positive number.
4. The leftmost question mark is the positive solution of 2 = 1 2, that is, = √12, and the rightmost question mark is the positive
solution of 2 = 3
2, that is, = 3 2. On the left side, we need | − 1| √12 − 1 ≈ 0292 (rounding down to be safe). On
the right side, we need | − 1|
3 2 − 1 ≈ 0224. The more restrictive of these two conditions must apply, so we choose
= 0224 (or any smaller positive number).
5. From the graph, we find that = tan = 08 when ≈ 0675, so
4
− 1 ≈ 0675 ⇒ 1 ≈
4 − 0675 ≈ 01106. Also, = tan = 12
when ≈ 0876, so 4 + 2 ≈ 0876 ⇒ 2 = 0876 − 4 ≈ 00906.
Thus, we choose = 00906 (or any smaller positive number) since this is
the smaller of 1 and 2.
6. From the graph, we find that = 2(2 + 4) = 03 when = 2 3, so
1 − 1 = 2
3 ⇒ 1 = 1 3. Also, = 2(2 + 4) = 05 when = 2, so
1 + 2 = 2 ⇒ 2 = 1. Thus, we choose = 1 3 (or any smaller positive
number) since this is the smaller of 1 and 2.
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7. From the graph with = 02, we find that = 3 − 3 + 4 = 58 when
≈ 19774, so 2 − 1 ≈ 19774 ⇒ 1 ≈ 00226. Also,
= 3 − 3 + 4 = 62 when ≈ 2022, so 2 + 2 ≈ 20219 ⇒
2 ≈ 00219. Thus, we choose = 00219 (or any smaller positive
number) since this is the smaller of 1 and 2.
For = 01, we get 1 ≈ 00112 and 2 ≈ 00110, so we choose
= 0011 (or any smaller positive number).
8. From the graph with = 05, we find that = (2 − 1) = 15 when
≈ −0303, so 1 ≈ 0303. Also, = (2 − 1) = 25 when
≈ 0215, so 2 ≈ 0215. Thus, we choose = 0215 (or any smaller
positive number) since this is the smaller of 1 and 2.
For = 01, we get 1 ≈ 0052 and 2 ≈ 0048, so we choose
= 0048 (or any smaller positive number).
9. (a)
The first graph of = 1
ln( − 1) shows a vertical asymptote at = 2. The second graph shows that = 100 when
≈ 201 (more accurately, 201005). Thus, we choose = 001 (or any smaller positive number).
(b) From part (a), we see that as gets closer to 2 from the right, increases without bound. In symbols,
lim
→2+
1
ln( − 1) = ∞.
10. We graph = csc2 and = 500. The graphs intersect at ≈ 3186, so
we choose = 3186 − ≈ 0044. Thus, if 0 | − | 0044, then
csc2 500. Similarly, for = 1000, we get = 3173 − ≈ 0031.
11. (a) = 2 and = 1000 cm2 ⇒ 2 = 1000 ⇒ 2 = 1000 ⇒ = 1000 ( 0) ≈ 178412 cm.
(b) | − 1000| ≤ 5 ⇒ −5 ≤ 2 − 1000 ≤ 5 ⇒ 1000 − 5 ≤ 2 ≤ 1000 + 5 ⇒
995 ≤ ≤ 1005 ⇒ 177966 ≤ ≤ 178858. 1000 − 995 ≈ 004466 and 1005 − 1000 ≈ 004455. So
if the machinist gets the radius within 00445 cm of 178412, the area will be within 5 cm2 of 1000.
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(c) is the radius, () is the area, is the target radius given in part (a), is the target area (1000), is the tolerance in the
area (5), and is the tolerance in the radius given in part (b).
12. (a) = 012 + 2155 + 20 and = 200 ⇒
012 + 2155 + 20 = 200 ⇒ [by the quadratic formula or
from the graph] ≈ 330 watts ( 0)
(b) From the graph, 199 ≤ ≤ 201 ⇒ 3289 3311.
(c) is the input power, () is the temperature, is the target input power given in part (a), is the target temperature (200),
is the tolerance in the temperature (1), and is the tolerance in the power input in watts indicated in part (b) (011 watts).
13. (a) |4 − 8| = 4 | − 2| 01 ⇔ | − 2| 01
4 , so = 041 = 0025.
(b) |4 − 8| = 4 | − 2| 001 ⇔ | − 2| 001
4 , so = 0401 = 00025.
14. |(5 − 7) − 3| = |5 − 10| = |5( − 2)| = 5 | − 2|. We must have |() − | , so 5 | − 2| ⇔
| − 2| 5. Thus, choose = 5. For = 01, = 002; for = 005, = 001; for = 001, = 0002.
15. Given 0, we need 0 such that if 0 | − 3| , then
(1 + 1 3) − 2 . But (1 + 1 3) − 2 ⇔ 1 3 − 1 ⇔
13
| − 3| ⇔ | − 3| 3. So if we choose = 3, then
0 | − 3| ⇒ (1 + 1 3) − 2 . Thus, lim
→3
(1 + 1 3) = 2 by
the definition of a limit.
16. Given 0, we need 0 such that if 0 | − 4| , then
|(2 − 5) − 3| . But |(2 − 5) − 3| ⇔ |2 − 8| ⇔
|2| | − 4| ⇔ | − 4| 2. So if we choose = 2, then
0 | − 4| ⇒ |(2 − 5) − 3| . Thus, lim
→4
(2 − 5) = 3 by the
definition of a limit.
17. Given 0, we need 0 such that if 0 | − (−3)| , then
|(1 − 4) − 13| . But |(1 − 4) − 13| ⇔
|−4 − 12| ⇔ |−4| | + 3| ⇔ | − (−3)| 4. So if
we choose = 4, then 0 | − (−3)| ⇒ |(1 − 4) − 13| .
Thus, lim
→−3
(1 − 4) = 13 by the definition of a limit.
x
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18. Given 0, we need 0 such that if 0 | − (−2)| , then
|(3 + 5) − (−1)| . But |(3 + 5) − (−1)| ⇔
|3 + 6| ⇔ |3| | + 2| ⇔ | + 2| 3. So if we choose
= 3, then 0 | + 2| ⇒ |(3 + 5) − (−1)| . Thus,
lim
→−2
(3 + 5) = −1 by the definition of a limit.
19. Given 0, we need 0 such that if 0 | − 1| , then
2 + 4
3
− 2
. But
2 + 4
3
− 2
⇔
4 − 4
3
⇔ 4 3 | − 1| ⇔ | − 1| 3 4. So if we choose = 3 4, then 0 | − 1| ⇒
2 + 4
3
− 2
. Thus, lim
→1
2 + 4
3
= 2 by the definition of a limit.
20. Given 0, we need 0 such that if 0 | − 10| , then 3 − 4 5 − (−5) . But 3 − 4 5 − (−5) ⇔
8 − 4
5
⇔ − 4 5 | − 10| ⇔ | − 10| 5 4. So if we choose = 5 4, then 0 | − 10| ⇒
3 − 4
5 − (−5) . Thus, lim
→10
(3 − 4 5) = −5 by the definition of a limit.
21. Given 0, we need 0 such that if 0 | − 4| , then
2 − 2 − 8
− 4 − 6
⇔
( − 4)( + 2)
− 4 − 6
⇔ | + 2 − 6| [ 6= 4] ⇔ | − 4| . So choose = . Then
0 | − 4| ⇒ | − 4| ⇒ | + 2 − 6| ⇒
( − 4)( + 2)
− 4 − 6
[ 6= 4] ⇒
2 − 2 − 8
− 4 − 6
. By the definition of a limit, lim
→4
2 − 2 − 8
− 4 = 6.
22. Given 0, we need 0 such that if 0 | + 15| , then
9 − 42
3 + 2
− 6
⇔
(3 + 2)(3 − 2)
3 + 2
− 6
⇔ |3 − 2 − 6| [ 6= −15] ⇔ |−2 − 3| ⇔ |−2| | + 15| ⇔
| + 15| 2. So choose = 2. Then 0 | + 15| ⇒ | + 15| 2 ⇒ |−2| | + 15| ⇒
|−2 − 3| ⇒ |3 − 2 − 6| ⇒
(3 + 2)(3 − 2)
3 + 2
− 6
[ 6= −15] ⇒
9 − 42
3 + 2
− 6
.
By the definition of a limit, lim
→−15
9 − 42
3 + 2
= 6.
23. Given 0, we need 0 such that if 0 | − | , then | − | . So = will work.
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24. Given 0, we need 0 such that if 0 | − | , then | − | . But | − | = 0, so this will be true no matter
what we pick.
25. Given 0, we need 0 such that if 0 | − 0| , then 2 − 0 ⇔ 2 ⇔ || √. Take = √.
Then 0 | − 0| ⇒ 2 − 0 . Thus, lim
→0
2 = 0 by the definition of a limit.
26. Given 0, we need 0 such that if 0 | − 0| , then 3 − 0 ⇔ ||3 ⇔ || √3 . Take = √3 .
Then 0 | − 0| ⇒ 3 − 0 3 = . Thus, lim
→0
3 = 0 by the definition of a limit.
27. Given 0, we need 0 such that if 0 | − 0| , then || − 0 . But || = ||. So this is true if we pick = .
Thus, lim
→0
|| = 0 by the definition of a limit.
28. Given 0, we need 0 such that if 0 − (−6) , then √8 6 + − 0 . But √8 6 + − 0 ⇔
√8 6 + ⇔ 6 + 8 ⇔ − (−6) 8. So if we choose = 8, then 0 − (−6) ⇒
√8 6 + − 0 . Thus, lim
→−6+
√8 6 + = 0 by the definition of a right-hand limit.
29. Given 0, we need 0 such that if 0 | − 2| , then 2 − 4 + 5 − 1 ⇔ 2 − 4 + 4 ⇔
( − 2)2 . So take = √. Then 0 | − 2| ⇔ | − 2| √ ⇔ ( − 2)2 . Thus,
lim
→2
2 − 4 + 5 = 1 by the definition of a limit.
30. Given 0, we need 0 such that if 0 | − 2| , then (2 + 2 − 7) − 1 . But (2 + 2 − 7) − 1 ⇔
2 + 2 − 8 ⇔ | + 4| | − 2| . Thus our goal is to make | − 2| small enough so that its product with | + 4|
is less than . Suppose we first require that | − 2| 1. Then −1 − 2 1 ⇒ 1 3 ⇒ 5 + 4 7 ⇒
| + 4| 7, and this gives us 7 | − 2| ⇒ | − 2| 7. Choose = min {1 7}. Then if 0 | − 2| , we
have | − 2| 7 and | + 4| 7, so (2 + 2 − 7) − 1 = |( + 4)( − 2)| = | + 4| | − 2| 7(7) = , as
desired. Thus, lim
→2
(2 + 2 − 7) = 1 by the definition of a limit.
31. Given 0, we need 0 such that if 0 | − (−2)| , then 2 − 1 − 3 or upon simplifying we need
2 − 4 whenever 0 | + 2| . Notice that if | + 2| 1, then −1 + 2 1 ⇒ −5 − 2 −3 ⇒
| − 2| 5. So take = min {5 1}. Then 0 | + 2| ⇒ | − 2| 5 and | + 2| 5, so
2 − 1 − 3 = |( + 2)( − 2)| = | + 2| | − 2| (5)(5) = . Thus, by the definition of a limit, lim
→−2
(2 − 1) = 3.
32. Given 0, we need 0 such that if 0 | − 2| , then 3 − 8 . Now 3 − 8 = ( − 2)2 + 2 + 4 .
If | − 2| 1, that is, 1 3, then 2 + 2 + 4 32 + 2(3) + 4 = 19 and so
3 − 8 = | − 2| 2 + 2 + 4 19 | − 2|. So if we take = min 1 19 , then 0 | − 2| ⇒
3 − 8 = | − 2| 2 + 2 + 4 19 · 19 = . Thus, by the definition of a limit, lim
→2
3 = 8.
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33. Given 0, we let = min 2 8. If 0 | − 3| , then | − 3| 2 ⇒ −2 − 3 2 ⇒
4 + 3 8 ⇒ | + 3| 8. Also | − 3| 8 , so 2 − 9 = | + 3| | − 3| 8 · 8 = . Thus, lim
→3
2 = 9.
34. From the figure, our choices for are 1 = 3 − √9 − and
2 = √9 + − 3. The largest possible choice for is the minimum
value of {1 2}; that is, = min{1 2} = 2 = √9 + − 3.
35. (a) The points of intersection in the graph are (1 26) and (2 34)
with 1 ≈ 0891 and 2 ≈ 1093. Thus, we can take to be the
smaller of 1 − 1 and 2 − 1. So = 2 − 1 ≈ 0093.
(b) Solving 3 + + 1 = 3 + gives us two nonreal complex roots and one real root, which is
() =
216 + 108 + 12√336 + 324 + 812 23 − 12
6216 + 108 + 12√336 + 324 + 812 13 . Thus, = () − 1.
(c) If = 04, then () ≈ 1093 272 342 and = () − 1 ≈ 0093, which agrees with our answer in part (a).
36. 1. Guessing a value for Let 0 be given. We have to find a number 0 such that
1
−
1 2
whenever
0 | − 2| . But
1
−
1 2
=
2 −
2
=
| − 2|
|2| . We find a positive constant such that |21| ⇒
| − 2|
|2| | − 2| and we can make | − 2| by taking | − 2| = . We restrict to lie in the interval
| − 2| 1 ⇒ 1 3 so 1 1
1 3
⇒
1 6
1
2
1 2
⇒
1
|2|
1 2
. So = 1
2
is suitable. Thus, we should
choose = min {1 2}.
2. Showing that works Given 0 we let = min {1 2}. If 0 | − 2| , then | − 2| 1 ⇒ 1 3 ⇒
1
|2|
1 2
(as in part 1). Also | − 2| 2, so
1
−
1 2
=
| − 2|
|2|
1 2
· 2 = . This shows that lim
→2
(1) = 1 2.
37. 1. Guessing a value for Given 0, we must find 0 such that |√ − √| whenever 0 | − | . But
|√ − √| = √| − + √| (from the hint). Now if we can find a positive constant such that √ + √ then
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| − |
√ + √
| − |
, and we take | − | . We can find this number by restricting to lie in some interval
centered at . If | − | 1 2, then − 1 2 − 1 2 ⇒ 1 2 3 2 ⇒ √ + √ 1 2 + √, and so
= 1 2 + √ is a suitable choice for the constant. So | − | 1 2 + √. This suggests that we let
= min 1 2 1 2 + √.
2. Showing that works Given 0, we let = min 1 2 1 2 + √. If 0 | − | , then
| − | 1 2 ⇒ √ + √ 1 2 + √ (as in part 1). Also | − | 1 2 + √, so
|√ − √ | = √| −+ √| 2 + 2 +√√ = . Therefore, lim → √ = √ by the definition of a limit.
38. Suppose that lim
→0
() = . Given = 1 2, there exists 0 such that 0 || ⇒ |() − | 1 2 ⇔
− 1
2 () + 1 2. For 0 , () = 1, so 1 + 1 2 ⇒ 1 2. For − 0, () = 0,
so − 1
2 0 ⇒ 1 2. This contradicts 1 2. Therefore, lim
→0
() does not exist.
39. Suppose that lim
→0
() = . Given = 1 2, there exists 0 such that 0 || ⇒ |() − | 1 2. Take any rational
number with 0 || . Then () = 0, so |0 − | 1 2, so ≤ || 1 2. Now take any irrational number with
0 || . Then () = 1, so |1 − | 1 2. Hence, 1 − 1 2, so 1 2. This contradicts 1 2, so lim
→0
() does not
exist.
40. First suppose that lim
→
() = . Then, given 0 there exists 0 so that 0 | − | ⇒ |() − | .
Then − ⇒ 0 | − | so |() − | . Thus, lim
→−
() = . Also + ⇒
0 | − | so |() − | . Hence, lim
→+
() = .
Now suppose lim
→−
() = = lim
→+
(). Let 0 be given. Since lim
→−
() = , there exists 1 0 so that
− 1 ⇒ |() − | . Since lim
→+
() = , there exists 2 0 so that + 2 ⇒
|() − | . Let be the smaller of 1 and 2. Then 0 | − | ⇒ − 1 or + 2 so
|() − | . Hence, lim
→
() = . So we have proved that lim
→
() = ⇔ lim
→−
() = = lim
→+
().
41. 1
( + 3)4 10,000 ⇔ ( + 3)4 10,1000 ⇔ | + 3| √4 101,000 ⇔ | − (−3)| 10 1
42. Given 0, we need 0 such that 0 | + 3| ⇒ 1( + 3)4 . Now 1
( + 3)4 ⇔
( + 3)4 1
⇔ | + 3| √41 . So take = √41 . Then 0 | + 3| = √41 ⇒ ( + 3) 1 4 , so
lim
→−3
1
( + 3)4 = ∞.
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43. Given 0 we need 0 so that ln whenever 0 ; that is, = ln whenever 0 . This
suggests that we take = . If 0 , then ln ln = . By the definition of a limit, lim
→0+
ln = −∞.
44. (a) Let be given. Since lim
→
() = ∞, there exists 1 0 such that 0 | − | 1 ⇒ () + 1 − . Since
lim
→
() = , there exists 2 0 such that 0 | − | 2 ⇒ |() − | 1 ⇒ () − 1. Let be the
smaller of 1 and 2. Then 0 | − | ⇒ () + () ( + 1 − ) + ( − 1) = . Thus,
lim
→
[() + ()] = ∞.
(b) Let 0 be given. Since lim
→
() = 0, there exists 1 0 such that 0 | − | 1 ⇒
|() − | 2 ⇒ () 2. Since lim
→
() = ∞, there exists 2 0 such that 0 | − | 2 ⇒
() 2. Let = min {1 2}. Then 0 | − | ⇒ () () 2
2
= , so lim
→
()() = ∞.
(c) Let 0 be given. Since lim
→
() = 0, there exists 1 0 such that 0 | − | 1 ⇒
|() − | −2 ⇒ () 2. Since lim
→
() = ∞, there exists 2 0 such that 0 | − | 2 ⇒
() 2. (Note that 0 and 0 ⇒ 2 0.) Let = min {1 2}. Then 0 | − | ⇒
() 2 ⇒ ()() 2
·
2
= , so lim
→
() () = −∞.
2.5 Continuity
1. From Definition 1, lim
→4
() = (4).
2. The graph of has no hole, jump, or vertical asymptote.
3. (a) is discontinuous at −4 since (−4) is not defined and at −2, 2, and 4 since the limit does not exist (the left and right
limits are not the same).
(b) is continuous from the left at −2 since lim
→−2−
() = (−2). is continuous from the right at 2 and 4 since
lim
→2+
() = (2) and lim
→4+
() = (4). It is continuous from neither side at −4 since (−4) is undefined.
4. From the graph of , we see that is continuous on the intervals [−3−2), (−2−1), (−1 0], (0 1), and (1 3].
5. The graph of = () must have a discontinuity at
= 2 and must show that lim
→2+
() = (2).
6. The graph of = () must have discontinuities
at = −1 and = 4. It must show that
lim
→−1−
() = (−1) and lim
→4+
() = (4).
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7. The graph of = () must have a removable
discontinuity (a hole) at = 3 and a jump discontinuity
at = 5.
8. The graph of = () must have a discontinuity
at = −2 with lim
→−2−
() 6= (−2) and
lim
→−2+
() 6= (−2). It must also show that
lim
→2−
() = (2) and lim
→2+
() 6= (2).
9. (a) The toll is $7 between 7:00 AM and 10:00 AM and between 4:00 PM and 7:00 PM.
(b) The function has jump discontinuities at = 7, 10, 16, and 19. Their
significance to someone who uses the road is that, because of the sudden jumps in
the toll, they may want to avoid the higher rates between = 7 and = 10 and
between = 16 and = 19 if feasible.
10. (a) Continuous; at the location in question, the temperature changes smoothly as time passes, without any instantaneous jumps
from one temperature to another.
(b) Continuous; the temperature at a specific time changes smoothly as the distance due west from New York City increases,
without any instantaneous jumps.
(c) Discontinuous; as the distance due west from New York City increases, the altitude above sea level may jump from one
height to another without going through all of the intermediate values— at a cliff, for example.
(d) Discontinuous; as the distance traveled increases, the cost of the ride jumps in small increments.
(e) Discontinuous; when the lights are switched on (or off ), the current suddenly changes between 0 and some nonzero value,
without passing through all of the intermediate values. This is debatable, though, depending on your definition of current.
11. lim
→−1
() = lim
→−1
+ 234 = lim →−1 + 2 lim →−1 34 = −1 + 2(−1)34 = (−3)4 = 81 = (−1).
By the definition of continuity, is continuous at = −1.
12. lim
→2
() = lim
→2
2 + 5
2 + 1
=
lim
→2
(2 + 5)
lim
→2
(2 + 1) =
lim
→2
2 + 5 lim
→2
2 lim
→2
+ lim
→2
1
=
22 + 5(2)
2(2) + 1 =
14
5
= (2).
By the definition of continuity, is continuous at = 2.
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13. lim
→1
() = lim
→1
2√32 + 1 = 2 lim
→1
√32 + 1 = 2lim →1(32 + 1) = 23 lim →1 2 + lim →1 1
= 23(1)2 + 1 = 2√4 = 4 = (1)
By the definition of continuity, is continuous at = 1.
14. lim
→2
() = lim
→2
34 − 5 + √3 2 + 4 = 3 lim
→2
4 − 5 lim
→2
+ 3 lim →2(2 + 4)
= 3(2)4 − 5(2) + √3 22 + 4 = 48 − 10 + 2 = 40 = (2)
By the definition of continuity, is continuous at = 2.
15. For 4, we have
lim
→
() = lim
→
( + √ − 4 ) = lim
→
+ lim
→
√ − 4 [Limit Law 1]
= + lim → − lim → 4 [8, 11, and 2]
= + √ − 4 [8 and 7]
= ()
So is continuous at = for every in (4 ∞). Also, lim
→4+
() = 4 = (4), so is continuous from the right at 4.
Thus, is continuous on [4 ∞).
16. For −2, we have
lim
→
() = lim
→
− 1
3 + 6
=
lim
→
( − 1)
lim
→
(3 + 6) [Limit Law 5]
=
lim
→
− lim
→
1
3 lim
→
+ lim
→
6
[2 1 and 3]
=
− 1
3 + 6
[8 and 7]
Thus, is continuous at = for every in (−∞ −2); that is, is continuous on (−∞ −2).
17. () = 1
+ 2
is discontinuous at = −2 because (−2) is undefined.
18. () =
1
+ 2
if 6= −2
1 if = −2
Here (−2) = 1, but lim
→−2−
() = −∞ and lim
→−2+
() = ∞,
so lim
→−2
() does not exist and is discontinuous at −2.
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19. () = 2+ 3 if if ≤ −−11
lim
→−1−
() = lim
→−1−
( + 3) = −1 + 3 = 2 and
lim
→−1+
() = lim
→−1+
2 = 2−1 = 1
2. Since the left-hand and the
right-hand limits of at −1 are not equal, lim
→−1
() does not exist, and
is discontinuous at −1.
20. () =
2 −
2 − 1 if 6= 1
1 if = 1
lim
→1
() = lim
→1
2 −
2 − 1 = lim →1 ( + 1)( ( −1) − 1) = lim →1 + 1 = 12,
but (1) = 1, so is discontinous at 1
21. () =
cos if 0
0 if = 0
1 − 2 if 0
lim
→0
() = 1, but (0) = 0 6= 1, so is discontinuous at 0.
22. () =
22 − 5 − 3
− 3
if 6= 3
6 if = 3
lim
→3
() = lim
→3
22 − 5 − 3
− 3
= lim
→3
(2 + 1)( − 3)
− 3
= lim
→3
(2 + 1) = 7,
but (3) = 6, so is discontinuous at 3.
23. () = 2 − − 2
− 2
=
( − 2)( + 1)
− 2
= + 1 for 6= 2. Since lim
→2
() = 2 + 1 = 3, define (2) = 3. Then is
continuous at 2.
24. () = 3 − 8
2 − 4 =
( − 2)(2 + 2 + 4)
( − 2)( + 2) =
2 + 2 + 4
+ 2
for 6= 2. Since lim
→2
() = 4 + 4 + 4
2 + 2
= 3, define (2) = 3.
Then is continuous at 2.
25. () = 22 − − 1
2 + 1 is a rational function, so it is continuous on its domain, (−∞ ∞), by Theorem 5(b).
26. () = 2 + 1
22 − − 1 =
2 + 1
(2 + 1)( − 1) is a rational function, so it is continuous on its domain,
−∞ − 1 2 ∪ − 1 2 1 ∪ (1 ∞), by Theorem 5(b).
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27. 3 − 2 = 0 ⇒ 3 = 2 ⇒ = √3 2, so () =
√3 − 2
3 − 2 has domain −∞ √3 2 ∪ √3 2 ∞. Now 3 − 2 is
continuous everywhere by Theorem 5(a) and √3 − 2 is continuous everywhere by Theorems 5(a), 7, and 9. Thus, is
continuous on its domain by part 5 of Theorem 4.
28. The domain of () = sin
2 + cos
is (−∞ ∞) since the denominator is never 0 [cos ≥ −1 ⇒ 2 + cos ≥ 1]. By
Theorems 7 and 9, sin and cos are continuous on R. By part 1 of Theorem 4, 2 + cos is continuous on R and by part 5
of Theorem 4, is continuous on R.
29. By Theorem 5(a), the polynomial 1 + 2 is continuous on R. By Theorem 7, the inverse trigonometric function arcsin is
continuous on its domain, [−1 1]. By Theorem 9, () = arcsin(1 + 2) is continuous on its domain, which is
{ | −1 ≤ 1 + 2 ≤ 1} = { | −2 ≤ 2 ≤ 0} = { | −1 ≤ ≤ 0} = [−1 0].
30. By Theorem 7, the trigonometric function tan is continuous on its domain, | 6= 2 + . By Theorems 5(a), 7, and 9,
the composite function √4 − 2 is continuous on its domain [−2 2]. By part 5 of Theorem 4, () = √tan 4 −2 is
continuous on its domain, (−2 −2) ∪ (−2 2) ∪ (2 2).
31. () = 1 + 1 = + 1 is defined when + 1 ≥ 0 ⇒ + 1 ≥ 0 and 0 or + 1 ≤ 0 and 0 ⇒ 0
or ≤ −1, so has domain (−∞ −1] ∪ (0 ∞). is the composite of a root function and a rational function, so it is
continuous at every number in its domain by Theorems 7 and 9.
32. By Theorems 7 and 9, the composite function −2 is continuous on R. By part 1 of Theorem 4, 1 + −2 is continuous on R.
By Theorem 7, the inverse trigonometric function tan−1 is continuous on its domain, R. By Theorem 9, the composite
function () = tan−11 + −2 is continuous on its domain, R.
33. The function = 1
1 + 1 is discontinuous at = 0 because the
left- and right-hand limits at = 0 are different.
34. The function = tan2 is discontinuous at = 2 + , where is
any integer. The function = lntan2 is also discontinuous
where tan2 is 0, that is, at = . So = lntan2 is
discontinuous at =
2 , any integer.
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35. Because is continuous on R and √20 − 2 is continuous on its domain, −√20 ≤ ≤ √20, the product
() = √20 − 2 is continuous on −√20 ≤ ≤ √20. The number 2 is in that domain, so is continuous at 2, and
lim
→2
() = (2) = 2√16 = 8.
36. Because is continuous on R, sin is continuous on R, and + sin is continuous on R, the composite function
() = sin( + sin ) is continuous on R, so lim
→
() = () = sin( + sin ) = sin = 0.
37. The function () = ln51 + − 2 is continuous throughout its domain because it is the composite of a logarithm function
and a rational function. For the domain of , we must have 5 − 2
1 +
0, so the numerator and denominator must have the
same sign, that is, the domain is (−∞ −√5 ] ∪ (−1 √5 ]. The number 1 is in that domain, so is continuous at 1, and
lim
→1
() = (1) = ln 5 − 1
1 + 1
= ln 2.
38. The function () = 3√2−2−4 is continuous throughout its domain because it is the composite of an exponential function,
a root function, and a polynomial. Its domain is
| 2 − 2 − 4 ≥ 0 = | 2 − 2 + 1 ≥ 5 = | ( − 1)2 ≥ 5
= | − 1| ≥ √5 = (−∞ 1 − √5 ] ∪ [1 + √5 ∞)
The number 4 is in that domain, so is continuous at 4, and lim
→4
() = (4) = 3√16−8−4 = 32 = 9.
39. () = 1 ln− 2 if if ≤ 1 1
By Theorem 5, since () equals the polynomial 1 − 2 on (−∞ 1], is continuous on (−∞ 1].
By Theorem 7, since () equals the logarithm function ln on (1 ∞), is continuous on (1 ∞).
At = 1, lim
→1−
() = lim
→1−
(1 − 2) = 1 − 12 = 0 and lim
→1+
() = lim
→1+
ln = ln 1 = 0. Thus, lim
→1
() exists and
equals 0. Also, (1) = 1 − 12 = 0. Thus, is continuous at = 1. We conclude that is continuous on (−∞ ∞).
40. () = sin cos if if ≥ 44
By Theorem 7, the trigonometric functions are continuous. Since () = sin on (−∞ 4) and () = cos on
(4 ∞), is continuous on (−∞ 4) ∪ (4 ∞) lim
→(4)−
() = lim
→(4)−
sin = sin
4 = 1√2 since the sine
function is continuous at 4 Similarly, lim
→(4)+
() = lim
→(4)+
cos = 1√2 by continuity of the cosine function
at 4. Thus, lim
→(4)
() exists and equals 1√2, which agrees with the value (4). Therefore, is continuous at 4,
so is continuous on (−∞ ∞).
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41. () =
2 if −1
if − 1 ≤ 1
1 if ≥ 1
is continuous on (−∞ −1), (−1 1), and (1 ∞), where it is a polynomial,
a polynomial, and a rational function, respectively.
Now lim
→−1−
() = lim
→−1−
2 = 1 and lim
→−1+
() = lim
→−1+
= −1,
so is discontinuous at −1. Since (−1) = −1, is continuous from the right at −1. Also, lim
→1−
() = lim
→1−
= 1 and
lim
→1+
() = lim
→1+
1
= 1 = (1), so is continuous at 1.
42. () =
2 if ≤ 1
3 − if 1 ≤ 4
√ if 4
is continuous on (−∞ 1), (1 4), and (4 ∞), where it is an exponential,
a polynomial, and a root function, respectively.
Now lim
→1−
() = lim
→1−
2 = 2 and lim
→1+
() = lim
→1+
(3 − ) = 2. Since (1) = 2 we have continuity at 1. Also,
lim
→4−
() = lim
→4−
(3 − ) = −1 = (4) and lim
→4+
() = lim
→4+
√ = 2, so is discontinuous at 4, but it is continuous
from the left at 4.
43. () =
+ 2 if 0
if 0 ≤ ≤ 1
2 − if 1
is continuous on (−∞ 0) and (1 ∞) since on each of these intervals
it is a polynomial; it is continuous on (0 1) since it is an exponential.
Now lim
→0−
() = lim
→0−
( + 2) = 2 and lim
→0+
() = lim
→0+
= 1, so is discontinuous at 0. Since (0) = 1, is
continuous from the right at 0. Also lim
→1−
() = lim
→1−
= and lim
→1+
() = lim
→1+
(2 − ) = 1, so is discontinuous
at 1. Since (1) = , is continuous from the left at 1.
44. By Theorem 5, each piece of is continuous on its domain. We need to check for continuity at = .
lim
→−
() = lim
→−
3 =
2 and →lim+ () = lim →+ 2 = 2 , so lim → () = 2 . Since () = 2 ,
is continuous at . Therefore, is a continuous function of .
45. () = 32−+ 2 if if ≥ 2 2
is continuous on (−∞ 2) and (2 ∞). Now lim
→2−
() = lim
→2−
2 + 2 = 4 + 4 and
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lim
→2+
() = lim
→2+
3 − = 8 − 2. So is continuous ⇔ 4 + 4 = 8 − 2 ⇔ 6 = 4 ⇔ = 2 3. Thus, for
to be continuous on (−∞ ∞), = 2 3.
46. () =
2 − 4
− 2
if 2
2 − + 3 if 2 ≤ 3
2 − + if ≥ 3
At = 2: lim
→2−
() = lim
→2−
2 − 4
− 2
= lim
→2−
( + 2)( − 2)
− 2
= lim
→2−
( + 2) = 2 + 2 = 4
lim
→2+
() = lim
→2+
(2 − + 3) = 4 − 2 + 3
We must have 4 − 2 + 3 = 4, or 4 − 2 = 1 (1).
At = 3: lim
→3−
() = lim
→3−
(2 − + 3) = 9 − 3 + 3
lim
→3+
() = lim
→3+
(2 − + ) = 6 − +
We must have 9 − 3 + 3 = 6 − + , or 10 − 4 = 3 (2).
Now solve the system of equations by adding −2 times equation (1) to equation (2).
−8 + 4 = −2
10 − 4 = 3
2 = 1
So = 1
2. Substituting 1 2 for in (1) gives us −2 = −1, so = 1 2 as well. Thus, for to be continuous on (−∞ ∞),
= = 1
2.
47. If and are continuous and (2) = 6, then lim
→2
[3() + () ()] = 36 ⇒
3 lim
→2
() + lim
→2
() · lim
→2
() = 36 ⇒ 3(2) + (2) · 6 = 36 ⇒ 9(2) = 36 ⇒ (2) = 4.
48. (a) () = 1
and () = 1
2 , so ( ◦ )() = (()) = (12) = 1 (12) = 2.
(b) The domain of ◦ is the set of numbers in the domain of (all nonzero reals) such that () is in the domain of (also
all nonzero reals). Thus, the domain is 6= 0 and 12 6= 0 = { | 6= 0} or (−∞ 0) ∪ (0 ∞). Since ◦ is
the composite of two rational functions, it is continuous throughout its domain; that is, everywhere except = 0.
49. (a) () = 4 − 1
− 1
=
(2 + 1)(2 − 1)
− 1
=
(2 + 1)( + 1)( − 1)
− 1
= (2 + 1)( + 1) [or 3 + 2 + + 1]
for 6= 1. The discontinuity is removable and () = 3 + 2 + + 1 agrees with for 6= 1 and is continuous on R.
(b) () = 3 − 2 − 2
− 2
=
(2 − − 2)
− 2
=
( − 2)( + 1)
− 2
= ( + 1) [or 2 + ] for 6= 2. The discontinuity
is removable and () = 2 + agrees with for 6= 2 and is continuous on R.
(c) lim
→−
() = lim
→−
[[sin ]] = lim
→−
0 = 0 and lim
→+
() = lim
→+
[[sin ]] = lim
→+
(−1) = −1, so lim
→
() does not
exist. The discontinuity at = is a jump discontinuity.
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50.
does not satisfy the conclusion of the
Intermediate Value Theorem.
does satisfy the conclusion of the
Intermediate Value Theorem.
51. () = 2 + 10 sin is continuous on the interval [31 32], (31) ≈ 957, and (32) ≈ 1030. Since 957 1000 1030,
there is a number c in (31 32) such that () = 1000 by the Intermediate Value Theorem. Note: There is also a number c in
(−32 −31) such that () = 1000
52. Suppose that (3) 6. By the Intermediate Value Theorem applied to the continuous function on the closed interval [2 3],
the fact that (2) = 8 6 and (3) 6 implies that there is a number in (2 3) such that () = 6. This contradicts the fact
that the only solutions of the equation () = 6 are = 1 and = 4. Hence, our supposition that (3) 6 was incorrect. It
follows that (3) ≥ 6. But (3) 6= 6 because the only solutions of () = 6 are = 1 and = 4. Therefore, (3) 6.
53. () = 4 + − 3 is continuous on the interval [1 2] (1) = −1, and (2) = 15. Since −1 0 15, there is a number
in (1 2) such that () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation 4 + − 3 = 0 in the
interval (1 2)
54. The equation ln = − √ is equivalent to the equation ln − + √ = 0. () = ln − + √ is continuous on the
interval [2 3], (2) = ln 2 − 2 + √2 ≈ 0107, and (3) = ln 3 − 3 + √3 ≈ −0169. Since (2) 0 (3), there is a
number in (2 3) such that () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation
ln − + √ = 0, or ln = − √, in the interval (2 3).
55. The equation = 3 − 2 is equivalent to the equation + 2 − 3 = 0. () = + 2 − 3 is continuous on the interval
[0 1], (0) = −2, and (1) = − 1 ≈ 172. Since −2 0 − 1, there is a number in (0 1) such that () = 0 by the
Intermediate Value Theorem. Thus, there is a root of the equation + 2 − 3 = 0, or = 3 − 2, in the interval (0 1).
56. The equation sin = 2 − is equivalent to the equation sin − 2 + = 0. () = sin − 2 + is continuous on the
interval [1 2] (1) = sin 1 ≈ 084, and (2) = sin 2 − 2 ≈ −109. Since sin 1 0 sin 2 − 2, there is a number in
(1 2) such that () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation sin − 2 + = 0, or
sin = 2 − , in the interval (1 2).
57. (a) () = cos − 3 is continuous on the interval [0 1], (0) = 1 0, and (1) = cos 1 − 1 ≈ −046 0. Since
1 0 −046, there is a number in (0 1) such that () = 0 by the Intermediate Value Theorem. Thus, there is a root
of the equation cos − 3 = 0, or cos = 3, in the interval (0 1).
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(b) (086) ≈ 0016 0 and (087) ≈ −0014 0, so there is a root between 086 and 087, that is, in the interval
(086 087).
58. (a) () = ln − 3 + 2 is continuous on the interval [1 2], (1) = −1 0, and (2) = ln 2 + 1 ≈ 17 0. Since
−1 0 17, there is a number in (1 2) such that () = 0 by the Intermediate Value Theorem. Thus, there is a root of
the equation ln − 3 + 2 = 0, or ln = 3 − 2, in the interval (1 2).
(b) (134) ≈ −003 0 and (135) ≈ 00001 0, so there is a root between 134 and 135 that is, in the
interval (134 135).
59. (a) Let () = 100−100 − 0012 Then (0) = 100 0 and
(100) = 100−1 − 100 ≈ −632 0. So by the Intermediate
Value Theorem, there is a number in (0 100) such that () = 0.
This implies that 100−100 = 0012.
(b) Using the intersect feature of the graphing device, we find that the
root of the equation is = 70347, correct to three decimal places.
60. (a) Let () = arctan + − 1. Then (0) = −1 0 and
(1) = 4 0. So by the Intermediate Value Theorem, there is a
number in (0 1) such that () = 0. This implies that
arctan = 1 − .
(b) Using the intersect feature of the graphing device, we find that the
root of the equation is = 0520, correct to three decimal places.
61. Let () = sin 3. Then is continuous on [1 2] since is the composite of the sine function and the cubing function, both
of which are continuous on R. The zeros of the sine are at , so we note that 0 1 3 2 2 8 3, and that the
pertinent cube roots are related by 1 3 3 2 [call this value ] 2. [By observation, we might notice that = √3 and
=
√3 2 are zeros of .]
Now (1) = sin 1 0, () = sin 3 2 = −1 0, and (2) = sin 8 0. Applying the Intermediate Value Theorem on
[1 ] and then on [ 2], we see there are numbers and in (1 ) and ( 2) such that () = () = 0. Thus, has at
least two -intercepts in (1 2).
62. Let () = 2 − 3 + 1. Then is continuous on (0 2] since is a rational function whose domain is (0 ∞). By
inspection, we see that 1 4 = 17 16 0, (1) = −1 0, and (2) = 3 2 0. Appling the Intermediate Value Theorem on
1 4 1 and then on [1 2], we see there are numbers and in 1 4 1 and (1 2) such that () = () = 0. Thus, has at
least two -intercepts in (0 2).
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63. (⇒) If is continuous at , then by Theorem 8 with () = + , we have
lim
→0
( + ) = lim →0 ( + ) = ().
(⇐) Let 0. Since lim
→0
( + ) = (), there exists 0 such that 0 || ⇒
|( + ) − ()| . So if 0 | − | , then |() − ()| = |( + ( − )) − ()| .
Thus, lim
→
() = () and so is continuous at .
64. lim
→0
sin( + ) = lim
→0
(sin cos + cos sin ) = lim
→0
(sin cos ) + lim
→0
(cos sin )
= lim →0 sin lim →0 cos + lim →0 cos lim →0 sin = (sin )(1) + (cos )(0) = sin
65. As in the previous exercise, we must show that lim
→0
cos( + ) = cos to prove that the cosine function is continuous.
lim
→0
cos( + ) = lim
→0
(cos cos − sin sin ) = lim
→0
(cos cos ) − lim
→0
(sin sin )
= lim →0 cos lim →0 cos − lim →0 sin lim →0 sin = (cos )(1) − (sin )(0) = cos
66. (a) Since is continuous at , lim
→
() = (). Thus, using the Constant Multiple Law of Limits, we have
lim
→
( )() = lim
→
() = lim
→
() = () = ( )(). Therefore, is continuous at .
(b) Since and are continuous at , lim
→
() = () and lim
→
() = (). Since () 6= 0, we can use the Quotient Law
of Limits: lim
→ () = lim → (()) = lim lim → ()
→
() =
()
() = (). Thus, is continuous at .
67. () = 0 1 if if is rational is irrational is continuous nowhere. For, given any number and any 0, the interval ( − + )
contains both infinitely many rational and infinitely many irrational numbers. Since () = 0 or 1, there are infinitely many
numbers with 0 | − | and |() − ()| = 1. Thus, lim
→
() 6= (). [In fact, lim
→
() does not even exist.]
68. () = 0 if if is rational is irrational is continuous at 0. To see why, note that − || ≤ () ≤ ||, so by the Squeeze Theorem
lim
→0
() = 0 = (0). But is continuous nowhere else. For if 6= 0 and 0, the interval ( − + ) contains both
infinitely many rational and infinitely many irrational numbers. Since () = 0 or , there are infinitely many numbers with
0 | − | and |() − ()| || 2. Thus, lim
→
() 6= ().
69. If there is such a number, it satisfies the equation 3 + 1 = ⇔ 3 − + 1 = 0. Let the left-hand side of this equation be
called (). Now (−2) = −5 0, and (−1) = 1 0. Note also that () is a polynomial, and thus continuous. So by the
Intermediate Value Theorem, there is a number between −2 and −1 such that () = 0, so that = 3 + 1.
70.
3 + 22 − 1 +
3 + − 2 = 0 ⇒ (3 + − 2) + (3 + 22 − 1) = 0. Let () denote the left side of the last
equation. Since is continuous on [−1 1], (−1) = −4 0, and (1) = 2 0, there exists a in (−1 1) such that
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() = 0 by the Intermediate Value Theorem. Note that the only root of either denominator that is in (−1 1) is
(−1 + √5 )2 = , but () = (3 √5 − 9)2 6= 0. Thus, is not a root of either denominator, so () = 0 ⇒
= is a root of the given equation.
71. () = 4 sin(1) is continuous on (−∞ 0) ∪ (0 ∞) since it is the product of a polynomial and a composite of a
trigonometric function and a rational function. Now since −1 ≤ sin(1) ≤ 1, we have −4 ≤ 4 sin(1) ≤ 4. Because
lim
→0
(−4) = 0 and lim
→0
4 = 0, the Squeeze Theorem gives us lim
→0
(4 sin(1)) = 0, which equals (0). Thus, is
continuous at 0 and, hence, on (−∞ ∞).
72. (a) lim
→0+
() = 0 and lim
→0−
() = 0, so lim
→0
() = 0, which is (0), and hence is continuous at = if = 0. For
0, lim
→
() = lim
→
= = (). For 0, lim
→
() = lim
→
(−) = − = (). Thus, is continuous at
= ; that is, continuous everywhere.
(b) Assume that is continuous on the interval . Then for ∈ , lim
→
|()| =
lim
→
()
= |()| by Theorem 8. (If is
an endpoint of , use the appropriate one-sided limit.) So || is continuous on .
(c) No, the converse is false. For example, the function () = −11 if if ≥ 00 is not continuous at = 0, but |()| = 1 is
continuous on R.
73. Define () to be the monk’s distance from the monastery, as a function of time (in hours), on the first day, and define ()
to be his distance from the monastery, as a function of time, on the second day. Let be the distance from the monastery to
the top of the mountain. From the given information we know that (0) = 0, (12) = , (0) = and (12) = 0. Now
consider the function − , which is clearly continuous. We calculate that ( − )(0) = − and ( − )(12) = .
So by the Intermediate Value Theorem, there must be some time 0 between 0 and 12 such that ( − )(0) = 0 ⇔
(0) = (0). So at time 0 after 7:00 AM, the monk will be at the same place on both days.
2.6 Limits at Infinity; Horizontal Asymptotes
1. (a) As becomes large, the values of () approach 5.
(b) As becomes large negative, the values of () approach 3.
2. (a) The graph of a function can intersect a
vertical asymptote in the sense that it can
meet but not cross it.
The graph of a function can intersect a horizontal asymptote.
It can even intersect its horizontal asymptote an infinite
number of times.
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(b) The graph of a function can have 0, 1, or 2 horizontal asymptotes. Representative examples are shown.
No horizontal asymptote One horizontal asymptote Two horizontal asymptotes
3. (a) lim
→∞
() = −2 (b) lim
→−∞
() = 2 (c) lim
→1
() = ∞
(d) lim
→3
() = −∞ (e) Vertical: = 1, = 3; horizontal: = −2, = 2
4. (a) lim
→∞
() = 2 (b) lim
→−∞
() = −1 (c) lim
→0
() = −∞
(d) lim
→2−
() = −∞ (e) lim
→2+
() = ∞ (f ) Vertical: = 0, = 2;
horizontal: = −1, = 2
5. lim
→0
() = −∞,
lim
→−∞
() = 5,
lim
→∞
() = −5
6. lim
→2
() = ∞, lim
→−2+
() = ∞,
lim
→−2−
() = −∞, lim
→−∞
() = 0,
lim
→∞
() = 0, (0) = 0
7. lim
→2
() = −∞, lim
→∞
() = ∞,
lim
→−∞
() = 0, lim
→0+
() = ∞,
lim
→0−
() = −∞
8. lim
→∞
() = 3,
lim
→2−
() = ∞,
lim
→2+
() = −∞,
is odd
9. (0) = 3, lim
→0−
() = 4,
lim
→0+
() = 2,
lim
→−∞
() = −∞, lim
→4−
() = −∞,
lim
→4+
() = ∞, lim
→∞
() = 3
10. lim
→3
() = −∞, lim
→∞
() = 2,
(0) = 0, is even
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11. If () = 22, then a calculator gives (0) = 0, (1) = 05, (2) = 1, (3) = 1125, (4) = 1, (5) = 078125,
(6) = 05625, (7) = 03828125, (8) = 025, (9) = 0158203125, (10) = 009765625, (20) ≈ 000038147,
(50) ≈ 22204 × 10−12, (100) ≈ 78886 × 10−27. It appears that lim
→∞
22 = 0.
12. (a) From a graph of () = (1 − 2) in a window of [0 10,000] by [0 02], we estimate that lim
→∞
() = 014
(to two decimal places.)
(b)
()
10,000 0135 308
100,000 0135 333
1,000,000 0135 335
From the table, we estimate that lim
→∞
() = 01353 (to four decimal places.)
13. lim
→∞
22 − 7
52 + − 3 = lim →∞ (5(22+2 − −7) 3) 2 2 [Divide both the numerator and denominator by (the highest power of that appears in the denominator)] 2
=
lim
→∞
(2 − 72)
lim
→∞
(5 + 1 − 32) [Limit Law 5]
=
lim
→∞
2 − lim
→∞
(72)
lim
→∞
5 + lim
→∞
(1) − lim
→∞
(32) [Limit Laws 1 and 2]
=
2 − 7 lim
→∞
(12)
5 + lim
→∞
(1) − 3 lim
→∞
(12) [Limit Laws 7 and 3]
=
2 − 7(0)
5 + 0 + 3(0) [Theorem 2.6.5]
=
2 5
14. lim
→∞ 93−3 + 8 5 +−34 = lim →∞ 93−3 + 8 5 +−34 [Limit Law 11]
= lim →∞ 9 + 8 33 − 52 −24 + 13 [Divide by 3]
=
lim
→∞
(9 + 82 − 43)
lim
→∞
(33 − 52 + 1) [Limit Law 5]
=
lim
→∞
9 + lim
→∞
(82) − lim
→∞
(43)
lim
→∞
(33) − lim
→∞
(52) + lim
→∞
1
[Limit Laws 1 and 2]
=
9 + 8 lim
→∞
(12) − 4 lim
→∞
(13)
3 lim
→∞
(13) − 5 lim
→∞
(12) + 1 [Limit Laws 7 and 3]
= 9 + 8(0) 3(0) − 5(0) + 1 − 4(0) [Theorem 2.6.5]
= 9 1 = √9 = 3
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15. lim
→∞
3 − 2
2 + 1
= lim
→∞
(3 − 2)
(2 + 1) = lim →∞ 3 2 + 1 − 2 = lim →∞ lim 3 − 2 lim →∞1
→∞
2 + lim
→∞
1 =
3 − 2(0)
2 + 0
=
3 2
16. lim
→∞
1 − 2
3 − + 1 = lim →∞ ((1 3 −−+ 1) 2) 3 3 = lim →∞ 1 −1 1 3 2−+ 1 13
=
lim
→∞
13 − lim
→∞
1
lim
→∞
1 − lim
→∞
12 + lim
→∞
13 =
0 − 0
1 − 0 + 0
= 0
17. lim
→−∞
− 2
2 + 1 = lim →−∞ ((2−+ 1) 2) 22 = lim →−∞ 1 1 + 1 − 22 2 = →−∞ limlim1 − 2 lim →−∞12
→−∞
1 + lim
→−∞
12 =
0 − 2(0)
1 + 0
= 0
18. lim
→−∞
43 + 62 − 2
23 − 4 + 5 = lim →−∞ (4 (233+ 6 − 42+ 5) − 2) 33 = lim →−∞ 24 + 6 − 4 2−+ 5 2 33 = 4 + 0 2 − 0 + 0 − 0 = 2
19. lim
→∞
√ + 2
2 − 2 = lim →∞ ((2 √−+22)) 22 = lim →∞ 1 2 32−+ 1 1 = 0 + 1 0 − 1 = −1
20. lim
→∞
− √
232 + 3 − 5 = lim →∞
− √ 32
(232 + 3 − 5) 32 = lim →∞ 2 + 31 1122−−5132 = 2 + 0 0 −−1 0 = −12
21. lim
→∞
(22 + 1)2
( − 1)2(2 + ) = lim →∞ [( −(21) 22+ 1) (2 +2 )] 44 = lim →∞ [(2 − 2[(2+ 1) 2 + 1) 2][( 22]2+ )2]
= lim
→∞
(2 + 12)2
(1 − 2 + 12)(1 + 1) =
(2 + 0)2
(1 − 0 + 0)(1 + 0) = 4
22. lim
→∞
2
√4 + 1 = lim →∞ √42+ 1 22 = lim →∞ (4 + 1) 1 4 [since 2 = √4 for 0]
= lim
→∞
1
1 + 14 =
1
√1 + 0 = 1
23. lim
→∞
√1 + 46
2 − 3 = lim →∞
√1 + 463
(2 − 3)3 =
lim
→∞
(1 + 46)6
lim
→∞
(23 − 1) [since 3 = √6 for 0]
=
lim
→∞
16 + 4
lim
→∞
(23) − lim
→∞
1
=
lim →∞(16) + lim →∞4
0 − 1
=
√0 + 4
−1
=
2 −
1
= −2
24. lim
→−∞
√1 + 46
2 − 3 = lim →−∞
√1 + 463
(2 − 3)3 =
lim
→−∞
−(1 + 46)6
lim
→−∞
(23 − 1) [since 3 = −√6 for 0]
=
lim
→−∞
−16 + 4
2 lim
→−∞
(13) − lim
→−∞
1
=
−→−∞ lim (16) + lim →−∞4
2(0) − 1
=
−√0 + 4
−1
=
−2
−1
= 2
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25. lim
→∞
√ + 32
4 − 1 = lim →∞
√ + 32
(4 − 1) =
lim
→∞
( + 32)2
lim
→∞
(4 − 1) [since = √2 for 0]
=
lim
→∞
1 + 3
lim
→∞
4 − lim
→∞
(1) =
lim →∞(1) + lim →∞3
4 − 0 =
√0 + 3
4 =
√3
4
26. lim
→∞
+ 32
4 − 1 = lim →∞ ((4 + 3 −1) 2) = lim →∞ 41 + 3 − 1
= ∞ since 1 + 3 → ∞ and 4 − 1 → 4 as → ∞.
27. lim
→∞
√92 + − 3 = lim
→∞
√92 + − 3√92 + + 3
√92 + + 3 = lim →∞
√92 + 2 − (3)2
√92 + + 3
= lim
→∞
92 + − 92
√92 + + 3 = lim →∞ √92 + + 3 · 11
= lim
→∞
922 + 2 + 3 = lim →∞ 9 + 11 + 3 = √9 + 3 1 = 3 + 3 1 = 1 6
28. lim
→−∞
√42 + 3 + 2 = lim
→−∞
√42 + 3 + 2 √√4 4 2 2 + 3 + 3 − − 2 2
= lim
→−∞
42 + 3 − (2)2
√42 + 3 − 2 = lim →−∞ √42 + 3 3 − 2
= lim
→−∞
3
√42 + 3 − 2 = lim →−∞ −4 + 3 3 − 2 [since = −√2 for 0]
=
3
−√4 + 0 − 2 = −
3 4
29. lim
→∞
√2 + − √2 + = lim
→∞
√2 + − √2 + √2 + + √2 +
√2 + + √2 +
= lim
→∞
(2 + ) − (2 + )
√2 + + √2 + = lim →∞ √2 + [(+−√)2]+ √2
= lim
→∞
−
1 + + 1 + =
−
√1 + 0 + √1 + 0 =
−
2
30. For 0, √2 + 1 √2 = . So as → ∞, we have √2 + 1 → ∞, that is, lim
→∞
√2 + 1 = ∞.
31. lim
→∞
4 − 32 +
3 − + 2 = lim →∞ ((4 3−−32+ 2) + )3 3 divide by the highest power of in the denominator = lim →∞ 1−−13 2+ 1 + 2 23 = ∞
since the numerator increases without bound and the denominator approaches 1 as → ∞.
32. lim
→∞
(− + 2 cos 3) does not exist. lim
→∞
− = 0, but lim
→∞
(2 cos 3) does not exist because the values of 2 cos 3
oscillate between the values of −2 and 2 infinitely often, so the given limit does not exist.
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33. lim
→−∞
(2 + 27) = lim
→−∞
7 15 + 2 [factor out the largest power of ] = −∞ because 7 → −∞ and
15 + 2 → 2 as → −∞.
Or: lim
→−∞
2 + 27 = lim
→−∞
2 1 + 25 = −∞.
34. lim
→−∞
1 + 6
4 + 1 = lim →−∞ (1 + (4 + 1) 6) 4 4 divide by the highest power of in the denominator = lim →−∞ 11 + 1 4 +42 = ∞
since the numerator increases without bound and the denominator approaches 1 as → −∞.
35. Let = . As → ∞, → ∞. lim
→∞
arctan() = lim
→∞
arctan =
2 by (3).
36. Divide numerator and denominator by 3: lim
→∞
3 − −3
3 + −3 = lim →∞ 1 1 +− − −6 6 = 11 + 0 − 0 = 1
37. lim
→∞
1 −
1 + 2
= lim
→∞
(1 − )
(1 + 2) = lim →∞ 1 1 − + 2 1 = 00 + 2 − 1 = −1 2
38. Since 0 ≤ sin2 ≤ 1, we have 0 ≤ sin2
2 + 1 ≤ 2 1+ 1. We know that lim →∞0 = 0 and lim →∞ 2 1+ 1 = 0, so by the Squeeze
Theorem, lim
→∞
sin2
2 + 1 = 0.
39. Since −1 ≤ cos ≤ 1 and −2 0, we have −−2 ≤ −2 cos ≤ −2. We know that lim
→∞
(−−2) = 0 and
lim
→∞
−2 = 0, so by the Squeeze Theorem, lim
→∞
(−2 cos ) = 0.
40. Let = ln . As → 0+, → −∞. lim
→0+
tan−1(ln ) = lim
→−∞
tan−1 = −
2 by (4).
41. lim
→∞
[ln(1 + 2) − ln(1 + )] = lim
→∞
ln 1 + 2
1 +
= ln lim →∞ 1 + 1 +2 = ln lim →∞ 1 1 ++ 1 = ∞, since the limit in
parentheses is ∞.
42. lim
→∞
[ln(2 + ) − ln(1 + )] = lim
→∞
ln 1 + 2 + = lim →∞ln 21 + 1 + 1 = ln 11 = ln 1 = 0
43. (a) (i) lim
→0+
() = lim
→0+
ln = 0 since → 0+ and ln → −∞ as → 0+.
(ii) lim
→1−
() = lim
→1−
ln = −∞ since → 1 and ln → 0− as → 1−.
(iii) lim
→1+
() = lim
→1+
ln = ∞ since → 1 and ln → 0+ as → 1+.
(b)
()
10,000 10857
100,000 86859
1,000,000 72,3824
It appears that lim
→∞
() = ∞.
(c)
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44. (a) lim
→∞
() = lim
→∞2 − ln1 = 0
since 2
→ 0 and 1
ln → 0 as → ∞.
(e)
(b) lim
→0+
() = lim
→0+ 2 − ln1 = ∞
since 2
→ ∞ and 1
ln → 0 as → 0+.
(c) lim
→1−
() = lim
→1− 2 − ln1 = ∞ since 2 → 2 and ln1 → −∞ as → 1−.
(d) lim
→1+
() = lim
→1+ 2 − ln1 = −∞ since 2 → 2 and ln1 → ∞ as → 1+.
45. (a)
From the graph of () = √2 + + 1 + , we
estimate the value of lim
→−∞
() to be −05.
(b)
()
−10,000 −0499 962 5
−100,000 −0499 996 2
−1,000,000 −0499 999 6
From the table, we estimate the limit to be −05.
(c) lim
→−∞
√2 + + 1 + = lim
→−∞
√2 + + 1 + √√ 2 2 + + + 1 + 1 − − = lim →−∞ √22+++ 1 + 1 −−2
= lim
→−∞
( + 1)(1)
√2 + + 1 − (1) = lim →−∞ −1 + (1 1 + (1 ) + (1 )2) − 1
=
1 + 0
−√1 + 0 + 0 − 1 = −
1 2
Note that for 0, we have √2 = || = −, so when we divide the radical by , with 0, we get
1
√2 + + 1 = −√12 √2 + + 1 = −1 + (1) + (12).
46. (a)
From the graph of
() = √32 + 8 + 6 − √32 + 3 + 1, we estimate
(to one decimal place) the value of lim
→∞
() to be 14.
(b)
()
10,000 1443 39
100,000 1443 38
1,000,000 1443 38
From the table, we estimate (to four decimal
places) the limit to be 14434.
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© Cengage Learning. All Rights Reserved.116 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
(c) lim
→∞
() = lim
→∞
√32 + 8 + 6 − √32 + 3 + 1 √32 + 8 + 6 + √32 + 3 + 1
√32 + 8 + 6 + √32 + 3 + 1
= lim
→∞
32 + 8 + 6 − 32 + 3 + 1
√32 + 8 + 6 + √32 + 3 + 1 = lim →∞ √32 + 8 + 6 + (5 + 5)(1 √32 + 3 ) + 1 (1)
= lim
→∞
5 + 5
3 + 8 + 62 + 3 + 3 + 12 =
5
√3 + √3 =
5
2 √3 =
5 √3
6
≈ 1443376
47. lim
→±∞
5 + 4
+ 3
= lim
→±∞
(5 + 4)
( + 3) = lim →±∞ 5 1 + 3 + 4 = 0 + 4 1 + 0 = 4, so
= 4 is a horizontal asymptote. = () = 5 + 4
+ 3
, so lim
→−3+
() = −∞
since 5 + 4 → −7 and + 3 → 0+ as → −3+. Thus, = −3 is a vertical
asymptote. The graph confirms our work.
48. lim
→±∞
22 + 1
32 + 2 − 1 = lim →±∞ (3(2 2 + 2 2 + 1) − 1)22
= lim
→±∞
2 + 12
3 + 2 − 12 =
2 3
so =
2 3
is a horizontal asymptote. = () = 22 + 1
32 + 2 − 1 =
22 + 1
(3 − 1)( + 1).
The denominator is zero when = 1
3 and −1, but the numerator is nonzero, so = 1 3 and = −1 are vertical asymptotes.
The graph confirms our work.
49. lim
→±∞
22 + − 1
2 + − 2 = lim →±∞
22 + − 1
2
2 + − 2
2
= lim
→±∞
2 + 1
−
1 2
1 + 1
−
2 2
=
lim
→±∞2 + 1 − 12
lim
→±∞1 + 1 − 22
=
lim
→±∞
2 + lim
→±∞
1
− lim
→±∞
1 2
lim
→±∞
1 + lim
→±∞
1
− 2 lim
→±∞
1 2
=
2 + 0 − 0
1 + 0 − 2(0) = 2, so = 2 is a horizontal asymptote.
= () = 22 + − 1
2 + − 2 =
(2 − 1)( + 1)
( + 2)( − 1) , so →− lim2− () = ∞,
lim
→−2+
() = −∞, lim
→1−
() = −∞, and lim
→1+
() = ∞. Thus, = −2
and = 1 are vertical asymptotes. The graph confirms our work.
50. lim
→±∞
1 + 4
2 − 4 = lim →±∞
1 + 4
4
2 − 4
4
= lim
→±∞
1 4
+ 1
1 2
− 1
=
lim
→±∞14 + 1
lim
→±∞12 − 1 =
lim
→±∞
1 4
+ lim
→±∞
1
lim
→±∞
1 2
− lim
→±∞
1
=
0 + 1
0 − 1
= −1, so = −1 is a horizontal asymptote.
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= () = 1 + 4
2 − 4 =
1 + 4
2(1 − 2) =
1 + 4
2(1 + )(1 − ). The denominator is
zero when = 0, −1, and 1, but the numerator is nonzero, so = 0, = −1, and
= 1 are vertical asymptotes. Notice that as → 0, the numerator and
denominator are both positive, so lim
→0
() = ∞. The graph confirms our work.
51. = () = 3 −
2 − 6 + 5 =
(2 − 1)
( − 1)( − 5) =
( + 1)( − 1)
( − 1)( − 5) =
( + 1)
− 5
= () for 6= 1.
The graph of is the same as the graph of with the exception of a hole in the
graph of at = 1. By long division, () = 2 +
− 5
= + 6 + 30
− 5
.
As → ±∞, () → ±∞, so there is no horizontal asymptote. The denominator
of is zero when = 5. lim
→5−
() = −∞ and lim
→5+
() = ∞, so = 5 is a
vertical asymptote. The graph confirms our work.
52. lim
→∞
2
− 5
= lim
→∞
2
− 5
·
1
1 = lim →∞ 1 − (5 2) = 1 −2 0 = 2, so = 2 is a horizontal asymptote.
lim
→−∞
2
− 5
=
2(0)
0 − 5
= 0, so = 0 is a horizontal asymptote. The denominator is zero (and the numerator isn’t)
when − 5 = 0 ⇒ = 5 ⇒ = ln 5.
lim
→(ln 5)+
2
− 5
= ∞ since the numerator approaches 10 and the denominator
approaches 0 through positive values as → (ln 5)+. Similarly,
lim
→(ln 5)−
2
− 5
= −∞. Thus, = ln 5 is a vertical asymptote. The graph
confirms our work.
53. From the graph, it appears = 1 is a horizontal asymptote.
lim
→±∞
33 + 5002
3 + 5002 + 100 + 2000 = lim →±∞
33 + 5002
3
3 + 5002 + 100 + 2000
3
= lim
→±∞
3 + (500)
1 + (500) + (1002) + (20003)
=
3 + 0
1 + 0 + 0 + 0
= 3, so = 3 is a horizontal asymptote.
The discrepancy can be explained by the choice of the viewing window. Try
[−100,000 100,000] by [−1 4] to get a graph that lends credibility to our
calculation that = 3 is a horizontal asymptote.
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54. (a)
From the graph, it appears at first that there is only one horizontal asymptote, at ≈ 0 and a vertical asymptote at
≈ 17. However, if we graph the function with a wider and shorter viewing rectangle, we see that in fact there seem to be
two horizontal asymptotes: one at ≈ 05 and one at ≈ −05. So we estimate that
lim
→∞
√22 + 1
3 − 5
≈ 05 and lim
→−∞
√22 + 1
3 − 5
≈ −05
(b) (1000) ≈ 04722 and (10,000) ≈ 04715, so we estimate that lim
→∞
√22 + 1
3 − 5
≈ 047.
(−1000) ≈ −04706 and (−10,000) ≈ −04713, so we estimate that lim
→−∞
√22 + 1
3 − 5
≈ −047.
(c) lim
→∞
√22 + 1
3 − 5
= lim
→∞
2 + 12
3 − 5 [since √2 = for 0] =
√2
3
≈ 0471404.
For 0, we have √2 = || = −, so when we divide the numerator by , with 0, we
get
1
√22 + 1 = −√12 √22 + 1 = −2 + 12. Therefore,
lim
→−∞
√22 + 1
3 − 5
= lim
→−∞
−2 + 12
3 − 5 = −
√2
3
≈ −0471404.
55. Divide the numerator and the denominator by the highest power of in ().
(a) If deg deg , then the numerator → 0 but the denominator doesn’t. So lim
→∞
[ ()()] = 0.
(b) If deg deg , then the numerator → ±∞ but the denominator doesn’t, so lim
→∞
[ ()()] = ±∞
(depending on the ratio of the leading coefficients of and ).
56.
(i) = 0 (ii) 0 ( odd) (iii) 0 ( even) (iv) 0 ( odd) (v) 0 ( even)
From these sketches we see that
(a) lim
→0+
=
1 if = 0
0 if 0
∞ if 0
(b) lim
→0−
=
1 if = 0
0 if 0
−∞ if 0, odd
∞ if 0, even
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(c) lim
→∞
=
1 if = 0
∞ if 0
0 if 0
(d) lim
→−∞
=
1 if = 0
−∞ if 0, odd
∞ if 0, even
0 if 0
57. Let’s look for a rational function.
(1) lim
→±∞
() = 0 ⇒ degree of numerator degree of denominator
(2) lim
→0
() = −∞ ⇒ there is a factor of 2 in the denominator (not just , since that would produce a sign
change at = 0), and the function is negative near = 0.
(3) lim
→3−
() = ∞ and lim
→3+
() = −∞ ⇒ vertical asymptote at = 3; there is a factor of ( − 3) in the
denominator.
(4) (2) = 0 ⇒ 2 is an -intercept; there is at least one factor of ( − 2) in the numerator.
Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us
() = 2 −
2( − 3) as one possibility.
58. Since the function has vertical asymptotes = 1 and = 3, the denominator of the rational function we are looking for must
have factors ( − 1) and ( − 3). Because the horizontal asymptote is = 1, the degree of the numerator must equal the
degree of the denominator, and the ratio of the leading coefficients must be 1. One possibility is () = 2
( − 1)( − 3).
59. (a) We must first find the function . Since has a vertical asymptote = 4 and -intercept = 1, − 4 is a factor of the
denominator and − 1 is a factor of the numerator. There is a removable discontinuity at = −1, so − (−1) = + 1 is
a factor of both the numerator and denominator. Thus, now looks like this: () = ( − 1)( + 1)
( − 4)( + 1) , where is still to
be determined. Then lim
→−1
() = lim
→−1
( − 1)( + 1)
( − 4)( + 1) = lim →−1 (−−41) = ((−−11−−4) 1) = 25, so 25 = 2, and
= 5. Thus () = 5( − 1)( + 1)
( − 4)( + 1) is a ratio of quadratic functions satisfying all the given conditions and
(0) = 5(−1)(1)
(−4)(1) =
5 4
.
(b) lim
→∞
() = 5 lim
→∞
2 − 1
2 − 3 − 4 = 5 lim →∞ (2(2)2−(3 2) − (1 2) −2(4 ) 2) = 51 −1 −0 −0 0 = 5(1) = 5
60. = () = 23 − 4 = 3(2 − ). The -intercept is (0) = 0. The
-intercepts are 0 and 2. There are sign changes at 0 and 2 (odd exponents on
and 2 − ). As → ∞, () → −∞ because 3 → ∞ and 2 − → −∞. As
→ −∞, () → −∞ because 3 → −∞ and 2 − → ∞. Note that the graph
of near = 0 flattens out (looks like = 3).
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61. = () = 4 − 6 = 4(1 − 2) = 4(1 + )(1 − ). The -intercept is
(0) = 0. The -intercepts are 0, −1, and 1 [found by solving () = 0 for ].
Since 4 0 for 6= 0, doesn’t change sign at = 0. The function does change
sign at = −1 and = 1. As → ±∞, () = 4(1 − 2) approaches −∞
because 4 → ∞ and (1 − 2) → −∞.
62. = () = 3( + 2)2( − 1). The -intercept is (0) = 0. The -intercepts
are 0, −2, and 1. There are sign changes at 0 and 1 (odd exponents on and
− 1). There is no sign change at −2. Also, () → ∞ as → ∞ because all
three factors are large. And () → ∞ as → −∞ because 3 → −∞,
( + 2)2 → ∞, and ( − 1) → −∞. Note that the graph of at = 0 flattens out
(looks like = −3).
63. = () = (3 − )(1 + )2(1 − )4. The -intercept is (0) = 3(1)2(1)4 = 3.
The -intercepts are 3, −1, and 1. There is a sign change at 3, but not at −1 and 1.
When is large positive, 3 − is negative and the other factors are positive, so
lim
→∞
() = −∞. When is large negative, 3 − is positive, so
lim
→−∞
() = ∞.
64. = () = 2(2 − 1)2( + 2) = 2( + 1)2( − 1)2( + 2). The
-intercept is (0) = 0. The -intercepts are 0, −1, 1 and −2. There is a sign
change at −2, but not at 0, −1, and 1. When is large positive, all the factors are
positive, so lim
→∞
() = ∞. When is large negative, only + 2 is negative, so
lim
→−∞
() = −∞.
65. (a) Since −1 ≤ sin ≤ 1 for all − 1
≤ sin
≤ 1
for 0. As → ∞, −1 → 0 and 1 → 0, so by the Squeeze
Theorem, (sin ) → 0. Thus, lim
→∞
sin
= 0.
(b) From part (a), the horizontal asymptote is = 0. The function
= (sin ) crosses the horizontal asymptote whenever sin = 0;
that is, at = for every integer . Thus, the graph crosses the
asymptote an infinite number of times.
66. (a) In both viewing rectangles,
lim
→∞
() = lim
→∞
() = ∞ and
lim
→−∞
() = lim
→−∞
() = −∞.
In the larger viewing rectangle, and
become less distinguishable.
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(b) lim
→∞
()
() = lim →∞ 35 −3553 + 2 = lim →∞1 − 53 · 12 + 23 · 14 = 1 − 5 3(0) + 2 3(0) = 1 ⇒
and have the same end behavior.
67. lim
→∞
5 √
√ − 1 ·
1√
1√ = lim →∞ 1 −5(1) = √15− 0 = 5 and
lim
→∞
10 − 21
2
·
1
1 = lim →∞ 10 − (21 2 ) = 102− 0 = 5. Since 102− 21 () √5√−1,
we have lim
→∞
() = 5 by the Squeeze Theorem.
68. (a) After minutes, 25 liters of brine with 30 g of salt per liter has been pumped into the tank, so it contains
(5000 + 25) liters of water and 25 · 30 = 750 grams of salt. Therefore, the salt concentration at time will be
() = 750
5000 + 25
=
30
200 +
gL
.
(b) lim
→∞
() = lim
→∞
30
200 +
= lim
→∞
30
200 + =
30
0 + 1
= 30. So the salt concentration approaches that of the brine
being pumped into the tank.
69. (a) lim
→∞
() = lim
→∞
∗1 − −∗ = ∗(1 − 0) = ∗
(b) We graph () = 1 − −98 and () = 099∗, or in this case,
() = 099. Using an intersect feature or zooming in on the point of
intersection, we find that ≈ 047 s.
70. (a) = −10 and = 01 intersect at 1 ≈ 2303.
If 1, then −10 01.
(b) −10 01 ⇒ −10 ln 01 ⇒
−10 ln 1
10 = −10 ln 10−1 = 10 ln 10 ≈ 2303
71. Let () = 32 + 1
22 + + 1 and () = |() − 15|. Note that
lim
→∞
() = 3 2 and lim
→∞
() = 0. We are interested in finding the
-value at which () 005. From the graph, we find that ≈ 14804,
so we choose = 15 (or any larger number).
72. We want to find a value of such that ⇒
1 − 3
√2 + 1 − (−3)
, or equivalently,
−3 − √1−2 + 1 3 −3 + . When = 01, we graph = () = √1−2 + 1 3 , = −31, and = −29. From the graph,
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we find that () = −29 at about = 11283, so we choose = 12 (or any larger number). Similarly for = 005, we find
that () = −295 at about = 21379, so we choose = 22 (or any larger number).
73. We want a value of such that ⇒
1 − 3
√2 + 1 − 3
, or equivalently, 3 − √1−2 + 1 3 3 + . When = 01,
we graph = () = √1−2 + 1 3 , = 31, and = 29. From the graph, we find that () = 31 at about = −8092, so we
choose = −9 (or any lesser number). Similarly for = 005, we find that () = 305 at about = −18338, so we
choose = −19 (or any lesser number).
74. We want to find a value of such that ⇒ √ ln 100.
We graph = () = √ ln and = 100. From the graph, we find
that () = 100 at about = 1382773, so we choose = 1383 (or
any larger number).
75. (a) 12 00001 ⇔ 2 100001 = 10 000 ⇔ 100 ( 0)
(b) If 0 is given, then 12 ⇔ 2 1 ⇔ 1√. Let = 1√.
Then ⇒ √1 ⇒ 12 − 0 = 12 , so lim →∞ 12 = 0.
76. (a) 1√ 00001 ⇔ √ 100001 = 104 ⇔ 108
(b) If 0 is given, then 1√ ⇔ √ 1 ⇔ 12. Let = 12.
Then ⇒ 1
2 ⇒
1 √
− 0
=
1 √
, so lim
→∞
1 √
= 0.
77. For 0, |1 − 0| = −1. If 0 is given, then −1 ⇔ −1.
Take = −1. Then ⇒ −1 ⇒ |(1) − 0| = −1 , so lim
→−∞
(1) = 0.
78. Given 0, we need 0 such that ⇒ 3 . Now 3 ⇔ √3 , so take = √3 . Then
= √3 ⇒ 3 , so lim
→∞
3 = ∞.
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79. Given 0, we need 0 such that ⇒ . Now ⇔ ln, so take
= max(1 ln). (This ensures that 0.) Then = max(1 ln) ⇒ max() ≥ ,
so lim
→∞
= ∞.
80. Definition Let be a function defined on some interval (−∞ ). Then lim
→−∞
() = −∞ means that for every negative
number there is a corresponding negative number such that () whenever . Now we use the definition to
prove that lim
→−∞
1 + 3 = −∞. Given a negative number , we need a negative number such that ⇒
1 + 3 . Now 1 + 3 ⇔ 3 − 1 ⇔ √3 − 1. Thus, we take = √3 − 1 and find that
⇒ 1 + 3 . This proves that lim
→−∞
1 + 3 = −∞.
81. (a) Suppose that lim
→∞
() = . Then for every 0 there is a corresponding positive number such that |() − |
whenever . If = 1, then ⇔ 0 1 1 ⇔ 0 1. Thus, for every 0 there is a
corresponding 0 (namely 1) such that |(1) − | whenever 0 . This proves that
lim
→0+
(1) = = lim
→∞
().
Now suppose that lim
→−∞
() = . Then for every 0 there is a corresponding negative number such that
|() − | whenever . If = 1, then ⇔ 1 1 0 ⇔ 1 0. Thus, for every
0 there is a corresponding 0 (namely −1) such that |(1) − | whenever − 0. This proves that
lim
→0−
(1) = = lim
→−∞
().
(b) lim
→0+
sin 1
= lim
→0+
sin 1
[let = ]
= lim
→∞
1
sin [part (a) with = 1]
= lim
→∞
sin
[let = ]
= 0 [by Exercise 65]
2.7 Derivatives and Rates of Change
1. (a) This is just the slope of the line through two points: = ∆
∆ =
() − (3)
− 3
.
(b) This is the limit of the slope of the secant line as approaches : = lim
→3
() − (3)
− 3
.
2. The curve looks more like a line as the viewing rectangle gets smaller.
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3. (a) (i) Using Definition 1 with () = 4 − 2 and (1 3),
= lim
→
() − ()
−
= lim
→1
(4 − 2) − 3
− 1
= lim
→1
−(2 − 4 + 3)
− 1
= lim
→1
−( − 1)( − 3)
− 1
= lim
→1
(3 − ) = 3 − 1 = 2
(ii) Using Equation 2 with () = 4 − 2 and (1 3),
= lim
→0
( + ) − ()
= lim →0 (1 + ) − (1) = lim →0 4(1 + ) −(1 + )2 − 3
= lim
→0
4 + 4 − 1 − 2 − 2 − 3
= lim →0 −2+ 2 = lim →0 (−+ 2) = lim →0(− + 2) = 2
(b) An equation of the tangent line is − () = 0()( − ) ⇒ − (1) = 0(1)( − 1) ⇒ − 3 = 2( − 1),
or = 2 + 1.
(c) The graph of = 2 + 1 is tangent to the graph of = 4 − 2 at the
point (1 3). Now zoom in toward the point (1 3) until the parabola and
the tangent line are indistiguishable.
4. (a) (i) Using Definition 1 with () = − 3 and (1 0),
= lim
→1
() − 0
− 1
= lim
→1
− 3
− 1
= lim
→1
(1 − 2)
− 1
= lim
→1
(1 + )(1 − )
− 1
= lim
→1
[−(1 + )] = −1(2) = −2
(ii) Using Equation 2 with () = − 3 and (1 0),
= lim
→0
( + ) − ()
= lim →0 (1 + ) − (1) = lim →0 (1 + ) − (1 + )3 − 0
= lim
→0
1 + − (1 + 3 + 32 + 3)
= lim →0 −3 − 32 − 2 = lim →0 (−2 −3 − 2)
= lim
→0
(−2 − 3 − 2) = −2
(b) An equation of the tangent line is − () = 0()( − ) ⇒ − (1) = 0(1)( − 1) ⇒ − 0 = −2( − 1),
or = −2 + 2.
(c) The graph of = −2 + 2 is tangent to the graph of = − 3 at the
point (1 0). Now zoom in toward the point (1 0) until the cubic and the
tangent line are indistinguishable.
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© Cengage Learning. All Rights Reserved.SECTION 2.7 DERIVATIVES AND RATES OF CHANGE ¤ 125
5. Using (1) with () = 4 − 32 and (2 −4) [we could also use (2)],
= lim
→
() − ()
−
= lim
→2
4 − 32− (−4)
− 2
= lim
→2
−32 + 4 + 4
− 2
= lim
→2
(−3 − 2)( − 2)
− 2
= lim
→2
(−3 − 2) = −3(2) − 2 = −8
Tangent line: − (−4) = −8( − 2) ⇔ + 4 = −8 + 16 ⇔ = −8 + 12.
6. Using (2) with () = 3 − 3 + 1 and (2 3),
= lim
→0
( + ) − ()
= lim →0 (2 + ) − (2) = lim →0 (2 + )3 − 3(2 + ) + 1 − 3
= lim
→0
8 + 12 + 62 + 3 − 6 − 3 − 2
= lim →0 9 + 62 + 3 = lim →0 (9 + 6 + 2)
= lim
→0
(9 + 6 + 2) = 9
Tangent line: − 3 = 9( − 2) ⇔ − 3 = 9 − 18 ⇔ = 9 − 15
7. Using (1), = lim
→1
√ − √1
− 1
= lim
→1
(√ − 1)(√ + 1)
( − 1)(√ + 1) = lim →1 ( − 1)( −√1 + 1) = lim →1 √1+ 1 = 12.
Tangent line: − 1 = 1 2( − 1) ⇔ = 1 2 + 1 2
8. Using (1) with () = 2 + 1
+ 2
and (1 1),
= lim
→
() − ()
−
= lim
→1
2 + 1
+ 2
− 1
− 1
= lim
→1
2 + 1 − ( + 2)
+ 2
− 1
= lim
→1
− 1
( − 1)( + 2)
= lim
→1
1
+ 2
=
1
1 + 2
=
1 3
Tangent line: − 1 = 1 3( − 1) ⇔ − 1 = 1 3 − 1 3 ⇔ = 1 3 + 2 3
9. (a) Using (2) with = () = 3 + 42 − 23,
= lim
→0
( + ) − ()
= lim →0 3 + 4( + )2 − 2( +)3 − (3 + 42 − 23)
= lim
→0
3 + 4(2 + 2 + 2) − 2(3 + 32 + 32 + 3) − 3 − 42 + 23
= lim
→0
3 + 42 + 8 + 42 − 23 − 62 − 62 − 23 − 3 − 42 + 23
= lim
→0
8 + 42 − 62 − 62 − 23
= lim →0 (8 + 4 − 62 − 6 − 22)
= lim
→0
(8 + 4 − 62 − 6 − 22) = 8 − 62
(b) At (1 5): = 8(1) − 6(1)2 = 2, so an equation of the tangent line
is − 5 = 2( − 1) ⇔ = 2 + 3.
At (2 3): = 8(2) − 6(2)2 = −8, so an equation of the tangent
line is − 3 = −8( − 2) ⇔ = −8 + 19.
(c)
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10. (a) Using (1),
= lim
→
1 √
−
1 √
−
= lim
→
√ − √
√
−
= lim
→
(√ − √ ) (√ + √ )
√ ( − ) (√ + √ ) = lim → √ ( −−) (√ + √ )
= lim
→
−1
√(√ + √ ) =
−1
√2 (2√ ) = −
1
232 or −
1 2
−32 [ 0]
(b) At (1 1): = − 1 2, so an equation of the tangent line
is − 1 = − 1 2( − 1) ⇔ = − 1 2 + 3 2.
At 4 1 2: = − 16 1 , so an equation of the tangent line
is − 1 2 = − 16 1 ( − 4) ⇔ = − 16 1 + 3 4.
(c)
11. (a) The particle is moving to the right when is increasing; that is, on the intervals (0 1) and (4 6). The particle is moving to
the left when is decreasing; that is, on the interval (2 3). The particle is standing still when is constant; that is, on the
intervals (1 2) and (3 4).
(b) The velocity of the particle is equal to the slope of the tangent line of the
graph. Note that there is no slope at the corner points on the graph. On the
interval (0 1) the slope is 3 − 0
1 − 0
= 3. On the interval (2 3), the slope is
1 − 3
3 − 2
= −2. On the interval (4 6), the slope is 3 − 1
6 − 4 = 1.
12. (a) Runner A runs the entire 100-meter race at the same velocity since the slope of the position function is constant.
Runner B starts the race at a slower velocity than runner A, but finishes the race at a faster velocity.
(b) The distance between the runners is the greatest at the time when the largest vertical line segment fits between the two
graphs—this appears to be somewhere between 9 and 10 seconds.
(c) The runners had the same velocity when the slopes of their respective position functions are equal—this also appears to be
at about 95 s. Note that the answers for parts (b) and (c) must be the same for these graphs because as soon as the velocity
for runner B overtakes the velocity for runner A, the distance between the runners starts to decrease.
13. Let () = 40 − 162.
(2) = lim
→2
() − (2)
− 2
= lim
→2
40 − 162 − 16
− 2
= lim
→2
−162 + 40 − 16
− 2
= lim
→2
−822 − 5 + 2
− 2
= lim
→2
−8( − 2)(2 − 1)
− 2
= −8 lim
→2
(2 − 1) = −8(3) = −24
Thus, the instantaneous velocity when = 2 is −24 fts.
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14. (a) Let () = 10 − 1862.
(1) = lim
→0
(1 + ) − (1)
= lim →0
10(1 + ) − 186(1 + )2 − (10 − 186)
= lim
→0
10 + 10 − 186(1 + 2 + 2) − 10 + 186
= lim
→0
10 + 10 − 186 − 372 − 1862 − 10 + 186
= lim
→0
628 − 1862
= lim →0(628 − 186) = 628
The velocity of the rock after one second is 628 ms.
(b) () = lim
→0
( + ) − ()
= lim →0
10( + ) − 186( + )2 − (10 − 1862)
= lim
→0
10 + 10 − 186(2 + 2 + 2) − 10 + 1862
= lim
→0
10 + 10 − 1862 − 372 − 1862 − 10 + 1862
= lim →0 10 − 372 − 1862
= lim
→0
(10 − 372 − 186)
= lim →0(10 − 372 − 186) = 10 − 372
The velocity of the rock when = is (10 − 372) ms
(c) The rock will hit the surface when = 0 ⇔ 10 − 1862 = 0 ⇔ (10 − 186) = 0 ⇔ = 0 or 186 = 10.
The rock hits the surface when = 10186 ≈ 54 s.
(d) The velocity of the rock when it hits the surface is 110 86 = 10 − 372 110 86 = 10 − 20 = −10 ms.
15. () = lim
→0
( + ) − ()
= lim →0
1
( + )2 −
1 2
= lim →0
2 − ( + )2
2( + )2
= lim →0 2 − (22(+ 2 + )+2 2)
= lim
→0
−(2 + 2)
2( + )2 = lim →0 −2(2 (++))2 = lim →0 −2(2 (++))2 = −2 ·22 = −32 ms
So (1) = −2
13 = −2 ms, (2) = −232 = −1 4ms, and (3) = −332 = −27 2 ms.
16. (a) The average velocity between times and + is
( + ) − ()
( + ) − =
12
( + )2 − 6( + ) + 23 − 1 22 − 6 + 23
=
12
2 + + 1
22 − 6 − 6 + 23 − 1 22 + 6 − 23
=
+ 1
22 − 6
=
+ 1 2 − 6
= + 1 2 − 6 fts
(i) [4 8]: = 4, = 8 − 4 = 4, so the average velocity is 4 + 1 2(4) − 6 = 0 fts.
(ii) [6 8]: = 6, = 8 − 6 = 2, so the average velocity is 6 + 1 2(2) − 6 = 1 fts.
(iii) [8 10]: = 8, = 10 − 8 = 2, so the average velocity is 8 + 1 2(2) − 6 = 3 fts.
(iv) [8 12]: = 8, = 12 − 8 = 4, so the average velocity is 8 + 1 2(4) − 6 = 4 fts.
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(b) () = lim
→0
( + ) − ()
= lim →0 + 1 2 − 6
= − 6, so (8) = 2 fts.
(c)
17. 0(0) is the only negative value. The slope at = 4 is smaller than the slope at = 2 and both are smaller than the slope
at = −2. Thus, 0(0) 0 0(4) 0(2) 0(−2).
18. (a) On [20 60]: (60) − (20)
60 − 20
=
700 − 300
40 =
400
40 = 10
(b) Pick any interval that has the same -value at its endpoints. [0 57] is such an interval since (0) = 600 and (57) = 600.
(c) On [40 60]: (60) − (40)
60 − 40 =
700 − 200
20
=
500
20
= 25
On [40 70]: (70) − (40)
70 − 40 =
900 − 200
30
=
700
30
= 23 1
3
Since 25 23 1
3, the average rate of change on [40 60] is larger.
(d) (40) − (10)
40 − 10 =
200 − 400
30
=
−200
30
= −6 2
3
This value represents the slope of the line segment from (10 (10)) to (40 (40)).
19. (a) The tangent line at = 50 appears to pass through the points (43 200) and (60 640), so
0(50) ≈ 640 − 200
60 − 43 =
440
17 ≈ 26.
(b) The tangent line at = 10 is steeper than the tangent line at = 30, so it is larger in magnitude, but less in numerical
value, that is, 0(10) 0(30).
(c) The slope of the tangent line at = 60, 0(60), is greater than the slope of the line through (40 (40)) and (80 (80)).
So yes, 0(60) (80) − (40)
80 − 40 .
20. Since (5) = −3, the point (5 −3) is on the graph of . Since 0(5) = 4, the slope of the tangent line at = 5 is 4.
Using the point-slope form of a line gives us − (−3) = 4( − 5), or = 4 − 23.
21. For the tangent line = 4 − 5: when = 2, = 4(2) − 5 = 3 and its slope is 4 (the coefficient of ). At the point of
tangency, these values are shared with the curve = (); that is, (2) = 3 and 0(2) = 4.
22. Since (4 3) is on = (), (4) = 3. The slope of the tangent line between (0 2) and (4 3) is 1 4, so 0(4) = 1 4.
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23. We begin by drawing a curve through the origin with a
slope of 3 to satisfy (0) = 0 and 0(0) = 3. Since
0(1) = 0, we will round off our figure so that there is
a horizontal tangent directly over = 1. Last, we
make sure that the curve has a slope of −1 as we pass
over = 2. Two of the many possibilities are shown.
24. We begin by drawing a curve through the origin with a slope of 1 to satisfy
(0) = 0 and 0(0) = 1. We round off our figure at = 1 to satisfy 0(1) = 0,
and then pass through (2 0) with slope −1 to satisfy (2) = 0 and 0(2) = −1.
We round the figure at = 3 to satisfy 0(3) = 0, and then pass through (4 0)
with slope 1 to satisfy (4) = 0 and 0(4) = 1 Finally we extend the curve on
both ends to satisfy lim
→∞
() = ∞ and lim
→−∞
() = −∞.
25. We begin by drawing a curve through (0 1) with a slope of 1 to satisfy (0) = 1
and 0(0) = 1. We round off our figure at = −2 to satisfy 0(−2) = 0. As
→ −5+, → ∞, so we draw a vertical asymptote at = −5. As → 5−,
→ 3, so we draw a dot at (5 3) [the dot could be open or closed].
26. We begin by drawing an odd function (symmetric with respect to the origin)
through the origin with slope −2 to satisfy 0(0) = −2. Now draw a curve starting
at = 1 and increasing without bound as → 2− since lim
→2−
() = ∞. Lastly,
reflect the last curve through the origin (rotate 180◦) since is an odd function.
27. Using (4) with () = 32 − 3 and = 1,
0(1) = lim
→0
(1 + ) − (1)
= lim →0 [3(1 + )2 −(1 + )3] − 2
= lim
→0
(3 + 6 + 32) − (1 + 3 + 32 + 3) − 2
= lim →0 3 − 3 = lim →0 (3 − 2)
= lim
→0
(3 − 2) = 3 − 0 = 3
Tangent line: − 2 = 3( − 1) ⇔ − 2 = 3 − 3 ⇔ = 3 − 1
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28. Using (5) with () = 4 − 2 and = 1,
0(1) = lim
→1
() − (1)
− 1
= lim
→1
(4 − 2) − (−1)
− 1
= lim
→1
4 − 1
− 1
= lim
→1
(2 + 1)(2 − 1)
− 1
= lim
→1
(2 + 1)( + 1)( − 1)
− 1
= lim
→1
[(2 + 1)( + 1)] = 2(2) = 4
Tangent line: − (−1) = 4( − 1) ⇔ + 1 = 4 − 4 ⇔ = 4 − 5
29. (a) Using (4) with () = 5(1 + 2) and the point (2 2), we have
0(2) = lim
→0
(2 + ) − (2)
= lim →0
5(2 + )
1 + (2 + )2 − 2
= lim
→0
5 + 10
2 + 4 + 5 − 2
= lim →0
5 + 10 − 2(2 + 4 + 5)
2 + 4 + 5
= lim
→0
−22 − 3
(2 + 4 + 5) = lim →0 ((2−+ 4 2 −+ 5) 3) = lim →0 2−+ 4 2 −+ 5 3 = −53
So an equation of the tangent line at (2 2) is − 2 = − 3 5( − 2) or = − 3 5 + 16 5 .
(b)
30. (a) Using (4) with () = 42 − 3, we have
0() = lim
→0
( + ) − ()
= lim →0 [4( + )2 − ( +)3] − (42 − 3)
= lim
→0
42 + 8 + 42 − (3 + 32 + 32 + 3) − 42 + 3
= lim
→0
8 + 42 − 32 − 32 − 3
= lim →0 (8 + 4 − 32 − 3 − 2)
= lim
→0
(8 + 4 − 32 − 3 − 2) = 8 − 32
At the point (2 8), 0(2) = 16 − 12 = 4, and an equation of the
tangent line is − 8 = 4( − 2), or = 4. At the point (3 9),
0(3) = 24 − 27 = −3, and an equation of the tangent line is
− 9 = −3( − 3), or = −3 + 18
(b)
31. Use (4) with () = 32 − 4 + 1.
0() = lim
→0
( + ) − ()
= lim →0 [3( + )2 − 4( + ) + 1] − (32 − 4 + 1)]
= lim
→0
32 + 6 + 32 − 4 − 4 + 1 − 32 + 4 − 1
= lim →0 6 + 32 − 4
= lim
→0
(6 + 3 − 4)
= lim →0(6 + 3 − 4) = 6 − 4
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32. Use (4) with () = 23 + .
0() = lim
→0
( + ) − ()
= lim →0 [2( + )3 + ( + )] − (23 + )
= lim
→0
23 + 62 + 62 + 23 + + − 23 −
= lim →0 62 + 62 + 23 +
= lim
→0
(62 + 6 + 22 + 1)
= lim →0(62 + 6 + 22 + 1) = 62 + 1
33. Use (4) with () = (2 + 1)( + 3).
0() = lim
→0
( + ) − ()
= lim →0
2( + ) + 1
( + ) + 3 −
2 + 1
+ 3
= lim
→0
(2 + 2 + 1)( + 3) − (2 + 1)( + + 3)
( + + 3)( + 3)
= lim
→0
(22 + 6 + 2 + 6 + + 3) − (22 + 2 + 6 + + + 3)
( + + 3)( + 3)
= lim
→0
5
( + + 3)( + 3) = lim →0 ( + + 3)( 5 + 3) = ( + 3) 5 2
34. Use (4) with () = −2 = 12.
0() = lim
→0
( + ) − ()
= lim →0
1
( + )2 −
1 2
= lim →0
2 − ( + )2
2( + )2
= lim
→0
2 − (2 + 2 + 2)
2( + )2 = lim →0 −22 ( +−)22 = lim →0 (−2(2+−))2
= lim
→0
−2 −
2( + )2 =
−2
2(2) =
−2
3
35. Use (4) with () = √1 − 2.
0() = lim
→0
( + ) − ()
= lim →0
1 − 2( + ) − √1 − 2
= lim
→0
1 − 2( + ) − √1 − 2
·
1 − 2( + ) + √1 − 2
1 − 2( + ) + √1 − 2
= lim
→0
1 − 2( + ) 2 − √1 − 2 2
1 − 2( + ) + √1 − 2 = lim →0 (11−−22(− +2)) + − (1√−1 −22)
= lim
→0
−2
1 − 2( + ) + √1 − 2 = lim →0 1 − 2( +−2) + √1 − 2
=
−2
√1 − 2 + √1 − 2 =
−2
2 √1 − 2 =
−1
√1 − 2
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36. Use (4) with () = √14− .
0() = lim
→0
( + ) − ()
= lim →0
4
1 − ( + ) −
4
√1 −
= 4 lim
→0
√1 − − √1 − −
√1 − − √1 −
= 4 lim →0
√1 − − √1 − −
√1 − − √1 −
= 4 lim
→0
√1 − − √1 − −
√1 − − √1 − ·
√1 − + √1 − −
√1 − + √1 − − = 4 lim →0 √1 − (−√1−√1)−2 −((√√11−−+−√1)2− − )
= 4 lim
→0
(1 − ) − (1 − − )
√1 − − √1 − (√1 − + √1 − − ) = 4 lim →0 √1 − − √1 − (√1 − + √1 − − )
= 4 lim
→0
1
√1 − − √1 − (√1 − + √1 − − ) = 4 · √1 − √1 − (√11 − + √1 − )
=
4
(1 − )(2√1 − ) =
2
(1 − )1(1 − )12 =
2
(1 − )32
37. By (4), lim
→0
√9 + − 3
= 0(9), where () = √ and = 9.
38. By (4), lim
→0
−2+ − −2
= 0(−2), where () = and = −2.
39. By Equation 5, lim
→2
6 − 64
− 2
= 0(2), where () = 6 and = 2.
40. By Equation 5, lim
→14
1
− 4
−
1 4
= 0(4), where () = 1
and = 1
4.
41. By (4), lim
→0
cos( + ) + 1
= 0(), where () = cos and = .
Or: By (4), lim
→0
cos( + ) + 1
= 0(0), where () = cos( + ) and = 0.
42. By Equation 5, lim
→6
sin − 1
2
−
6
= 0 6 , where () = sin and = 6 .
43. (4) = 0(4) = lim
→0
(4 + ) − (4)
= lim →0
80(4 + ) − 6(4 + )2 − 80(4) − 6(4)2
= lim
→0
(320 + 80 − 96 − 48 − 62) − (320 − 96)
= lim →0 32 − 62
= lim
→0
(32 − 6)
= lim →0(32 − 6) = 32 m/s
The speed when = 4 is |32| = 32 ms.
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44. (4) = 0(4) = lim
→0
(4 + ) − (4)
= lim →0
10 + 4 +45 + 1 − 10 + 4 + 1 45
= lim →0
45
5 + − 9
= lim
→0
45 − 9(5 + )
(5 + ) = lim →0 (5 + −9) = lim →0 5 + −9 = −95 m/s.
The speed when = 4 is − 9 5 = 9 5 ms.
45. The sketch shows the graph for a room temperature of 72◦ and a refrigerator
temperature of 38◦. The initial rate of change is greater in magnitude than the
rate of change after an hour.
46. The slope of the tangent (that is, the rate of change of temperature with respect
to time) at = 1 h seems to be about 75 − 168
132 − 0
≈ −07 ◦Fmin.
47. (a) (i) [10 20]: (2) − (1)
2 − 1
=
018 − 033
1
= −015 mg/mL
h
(ii) [15 20]: (2) − (15)
2 − 15
=
018 − 024
05
=
−006
05
= −012 mg/mL
h
(iii) [20 25]: (25) − (2)
25 − 2
=
012 − 018
05
=
−006
05
= −012 mg/mL
h
(iv) [20 30]: (3) − (2)
3 − 2
=
007 − 018
1
= −011 mg/mL
h
(b) We estimate the instantaneous rate of change at = 2 by averaging the average rates of change for [15 20] and [20 25]:
−012 + (−012)
2
= −012 mg/mL
h . After 2 hours, the BAC is decreasing at a rate of 012 (mgmL)h.
48. (a) (i) [2006 2008]: (2008) − (2006)
2008 − 2006
=
16,680 − 12,440
2
=
4240
2
= 2120 locationsyear
(ii) [2008 2010]: (2010) − (2008)
2010 − 2008
=
16,858 − 16,680
2
=
178
2
= 89 locationsyear.
The rate of growth decreased over the period from 2006 to 2010.
(b) [2010 2012]: (2012) − (2010)
2012 − 2010
=
18,066 − 16,858
2
=
1208
2
= 604 locationsyear.
Using that value and the value from part (a)(ii), we have 89 + 604
2
=
693
2
= 3465 locationsyear.
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(c) The tangent segment has endpoints (2008 16,250) and (2012 17,500).
An estimate of the instantaneous rate of growth in 2010 is
17,500 − 16,250
2012 − 2008
=
1250
4 = 3125 locations/year.
49. (a) [1990 2005]: 84,077 − 66,533
2005 − 1990
=
17,544
15
= 11696 thousands of barrels per day per year. This means that oil
consumption rose by an average of 11696 thousands of barrels per day each year from 1990 to 2005.
(b) [1995 2000]: 76,784 − 70,099
2000 − 1995
=
6685
5
= 1337
[2000 2005]: 84,077 − 76,784
2005 − 2000
=
7293
5
= 14586
An estimate of the instantaneous rate of change in 2000 is 1 2 (1337 + 14586) = 13978 thousands of barrels
per day per year.
50. (a) (i) [4 11]: (11) − (4)
11 − 4 =
94 − 53
7 =
−436
7 ≈ −623 RNA copies day mL
(ii) [8 11]: (11) − (8)
11 − 8
=
94 − 18
3
=
−86
3
≈ −287 RNA copiesmL
day
(iii) [11 15]: (15) − (11)
15 − 11
=
52 − 94
4 =
−42
4 = −105 RNA copies day mL
(iv) [11 22]: (22) − (11)
22 − 11
=
36 − 94
11
=
−58
11
≈ −053 RNA copiesmL
day
(b) An estimate of 0(11) is the average of the answers from part (a)(ii) and (iii).
0(11) ≈ 1 2 [−287 + (−105)] = −196 RNA copiesmL
day .
0(11) measures the instantaneous rate of change of patient 303’s viral load 11 days after ABT-538 treatment began.
51. (a) (i) ∆
∆ =
(105) − (100)
105 − 100
=
660125 − 6500
5
= $2025unit.
(ii) ∆
∆ =
(101) − (100)
101 − 100
=
652005 − 6500
1
= $2005unit.
(b) (100 + ) − (100)
=
5000 + 10(100 + ) + 005(100 + )2 − 6500
=
20 + 0052
= 20 + 005, 6= 0
So the instantaneous rate of change is lim
→0
(100 + ) − (100)
= lim →0(20 + 005) = $20unit.
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52. ∆ = ( + ) − () = 100,000 1 − +602 − 100,000 1 − 60 2
= 100,000 1 − 30 + + (3600 + )2 − 1 − 30 + 3600 2 = 100,000−30 + 3600 2 + 3600 2
=
100,000
3600
(−120 + 2 + ) = 250
9
(−120 + 2 + )
Dividing ∆ by and then letting → 0, we see that the instantaneous rate of change is 500 9 ( − 60) galmin.
Flow rate (galmin) Water remaining () (gal)
0 −33333 100 000
10 −27777 69 4444
20 −22222 44 4444
30 −16666 25 000
40 −11111 11 1111
50 − 5555 2 7777
60 0 0
The magnitude of the flow rate is greatest at the beginning and gradually decreases to 0.
53. (a) 0() is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are
dollars per ounce.
(b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is $17ounce. So the cost
of producing the 800th (or 801st) ounce is about $17.
(c) In the short term, the values of 0() will decrease because more efficient use is made of start-up costs as increases. But
eventually 0() might increase due to large-scale operations.
54. (a) 0(5) is the rate of growth of the bacteria population when = 5 hours. Its units are bacteria per hour.
(b) With unlimited space and nutrients, 0 should increase as increases; so 0(5) 0(10). If the supply of nutrients is
limited, the growth rate slows down at some point in time, and the opposite may be true.
55. (a) 0(58) is the rate at which the daily heating cost changes with respect to temperature when the outside temperature is
58 ◦F. The units are dollars ◦F.
(b) If the outside temperature increases, the building should require less heating, so we would expect 0(58) to be negative.
56. (a) 0(8) is the rate of change of the quantity of coffee sold with respect to the price per pound when the price is $8 per pound.
The units for 0(8) are pounds(dollarspound).
(b) 0(8) is negative since the quantity of coffee sold will decrease as the price charged for it increases. People are generally
less willing to buy a product when its price increases.
57. (a) 0() is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mgL)◦C.
(b) For = 16◦C, it appears that the tangent line to the curve goes through the points (0 14) and (32 6). So
0(16) ≈ 6 − 14
32 − 0
= −
8
32
= −025 (mgL)◦C. This means that as the temperature increases past 16◦C, the oxygen
solubility is decreasing at a rate of 025 (mgL)◦C.
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58. (a) 0() is the rate of change of the maximum sustainable speed of Coho salmon with respect to the temperature. Its units
are (cms)◦C.
(b) For = 15◦C, it appears the tangent line to the curve goes through the points (10 25) and (20 32). So
0(15) ≈ 32 − 25
20 − 10
= 07 (cms)◦C. This tells us that at = 15◦C, the maximum sustainable speed of Coho salmon is
changing at a rate of 0.7 (cms)◦C. In a similar fashion for = 25◦C, we can use the points (20 35) and (25 25) to
obtain 0(25) ≈ 25 − 35
25 − 20
= −2 (cms)◦C. As it gets warmer than 20◦C, the maximum sustainable speed decreases
rapidly.
59. Since () = sin(1) when 6= 0 and (0) = 0, we have
0(0) = lim
→0
(0 + ) − (0)
= lim →0 sin(1) − 0 = lim →0 sin(1). This limit does not exist since sin(1) takes the
values −1 and 1 on any interval containing 0. (Compare with Example 2.2.4.)
60. Since () = 2 sin(1) when 6= 0 and (0) = 0, we have
0(0) = lim
→0
(0 + ) − (0)
= lim →0 2 sin(1) − 0 = lim →0 sin(1). Since −1 ≤ sin 1 ≤ 1, we have
− || ≤ || sin 1
≤ || ⇒ − || ≤ sin 1 ≤ ||. Because lim →0 (− ||) = 0 and lim →0 || = 0, we know that
lim
→0 sin 1 = 0 by the Squeeze Theorem. Thus, 0(0) = 0.
61. (a) The slope at the origin appears to be 1.
(b) The slope at the origin still appears to be 1.
(c) Yes, the slope at the origin now appears to be 0.
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2.8 The Derivative as a Function
1. It appears that is an odd function, so 0 will be an even function—that
is, 0(−) = 0().
(a) 0(−3) ≈ −02
(b) 0(−2) ≈ 0 (c) 0(−1) ≈ 1 (d) 0(0) ≈ 2
(e) 0(1) ≈ 1 (f) 0(2) ≈ 0 (g) 0(3) ≈ −02
2. Your answers may vary depending on your estimates.
(a) Note: By estimating the slopes of tangent lines on the
graph of , it appears that 0(0) ≈ 6.
(b) 0(1) ≈ 0
(c) 0(2) ≈ −15 (d) 0(3) ≈ −13 (e) 0(4) ≈ −08
(f) 0(5) ≈ −03 (g) 0(6) ≈ 0 (h) 0(7) ≈ 02
3. (a)0 = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then
negative again. The actual function values in graph II follow the same pattern.
(b)0 = IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly
become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.
(c)0 = I, since the slopes of the tangents to graph (c) are negative for 0 and positive for 0, as are the function values of
graph I.
(d)0 = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then
positive, then 0, then negative again, and the function values in graph III follow the same pattern.
Hints for Exercises 4 –11: First plot -intercepts on the graph of 0 for any horizontal tangents on the graph of . Look for any corners on the graph
of — there will be a discontinuity on the graph of 0. On any interval where has a tangent with positive (or negative) slope, the graph of 0 will be
positive (or negative). If the graph of the function is linear, the graph of 0 will be a horizontal line.
4. 5.
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6. 7.
9. 9.
10. 11.
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12. The slopes of the tangent lines on the graph of = () are always
positive, so the -values of = 0() are always positive. These values start
out relatively small and keep increasing, reaching a maximum at about
= 6. Then the -values of = 0() decrease and get close to zero. The
graph of 0 tells us that the yeast culture grows most rapidly after 6 hours
and then the growth rate declines.
13. (a) 0() is the instantaneous rate of change of percentage
of full capacity with respect to elapsed time in hours.
(b) The graph of 0() tells us that the rate of change of
percentage of full capacity is decreasing and
approaching 0.
14. (a) 0() is the instantaneous rate of change of fuel
economy with respect to speed.
(b) Graphs will vary depending on estimates of 0, but
will change from positive to negative at about = 50.
(c) To save on gas, drive at the speed where is a
maximum and 0 is 0, which is about 50 mi h.
15. It appears that there are horizontal tangents on the graph of for = 1963
and = 1971. Thus, there are zeros for those values of on the graph of
0. The derivative is negative for the years 1963 to 1971.
16. See Figure 3.3.1.
17.
The slope at 0 appears to be 1 and the slope at 1 appears
to be 27. As decreases, the slope gets closer to 0. Since
the graphs are so similar, we might guess that 0() = .
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18.
As increases toward 1, 0() decreases from very large
numbers to 1. As becomes large, 0() gets closer to 0.
As a guess, 0() = 12 or 0() = 1 makes sense.
19. (a) By zooming in, we estimate that 0(0) = 0, 0 1 2 = 1, 0(1) = 2,
and 0(2) = 4.
(b) By symmetry, 0(−) = − 0(). So 0− 1 2 = −1, 0(−1) = −2,
and 0(−2) = −4.
(c) It appears that 0() is twice the value of , so we guess that 0() = 2.
(d) 0() = lim
→0
( + ) − ()
= lim →0 ( + )2 − 2
= lim
→0
2 + 2 + 2 − 2
= lim →0 2+ 2 = lim →0 (2+ ) = lim →0(2 + ) = 2
20. (a) By zooming in, we estimate that 0(0) = 0, 0 1 2 ≈ 075, 0(1) ≈ 3, 0(2) ≈ 12, and 0(3) ≈ 27.
(b) By symmetry, 0(−) = 0(). So 0− 1 2 ≈ 075, 0(−1) ≈ 3, 0(−2) ≈ 12, and 0(−3) ≈ 27.
(c) (d) Since 0(0) = 0, it appears that 0 may have the form 0() = 2.
Using 0(1) = 3, we have = 3, so 0() = 32.
(e) 0() = lim
→0
( + ) − ()
= lim →0 ( + )3 − 3 = lim →0 (3 + 32 + 32 + 3) − 3
= lim
→0
32 + 32 + 3
= lim →0 (32 + 3 + 2) = lim →0(32 + 3 + 2) = 32
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21. 0() = lim
→0
( + ) − ()
= lim →0 [3( + ) −8] − (3 − 8) = lim →0 3 + 3 −8 − 3 + 8
= lim
→0
3
= lim →0 3 = 3
Domain of = domain of 0 = R.
22. 0() = lim
→0
( + ) − ()
= lim →0 [( + ) +] − ( + ) = lim →0 + + − −
= lim
→0
= lim →0 =
Domain of = domain of 0 = R.
23. 0() = lim
→0
( + ) − ()
= lim →0
25( + )2 + 6( + ) − 252 + 6
= lim
→0
25(2 + 2 + 2) + 6 + 6 − 252 − 6
= lim →0 252 + 5 + 252 + 6 − 252
= lim
→0
5 + 252 + 6
= lim →0 (5 + 25 + 6) = lim →0 (5 + 25 + 6)
= 5 + 6
Domain of = domain of 0 = R.
24. 0() = lim
→0
( + ) − ()
= lim →0
4 + 8( + ) − 5( + )2 − (4 + 8 − 52)
= lim
→0
4 + 8 + 8 − 5(2 + 2 + 2) − 4 − 8 + 52
= lim →0 8 − 52 − 10 − 52 + 52
= lim
→0
8 − 10 − 52
= lim →0 (8 − 10 − 5) = lim →0 (8 − 10 − 5)
= 8 − 10
Domain of = domain of 0 = R.
25. 0() = lim
→0
( + ) − ()
= lim →0 [( + )2 − 2( +)3] − (2 − 23)
= lim
→0
2 + 2 + 2 − 23 − 62 − 62 − 23 − 2 + 23
= lim
→0
2 + 2 − 62 − 62 − 23
= lim →0 (2 + − 62 − 6 − 22)
= lim
→0
(2 + − 62 − 6 − 22) = 2 − 62
Domain of = domain of 0 = R.
26. 0() = lim
→0
( + ) − ()
= lim →0
1
√ + −
1 √
= lim →0
√ − √ +
√ + √
= lim →0 √√− +√+√ · √√ + + √ √ + +
= lim
→0
− ( + )
√ + √ √ + √ + = lim →0 √ + √−√ + √ + = lim →0 √ + √ −√1 + √ +
=
−1
√ √ √ + √ =
−1
2 √ = −
1
232
Domain of = domain of 0 = (0 ∞).
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27. 0() = lim
→0
( + ) − ()
= lim →0
9 − ( + ) − √9 −
9 9 − − ( ( + + ) + ) + √ √9 9 − −
= lim
→0
[9 − ( + )] − (9 − )
9 − ( + ) + √9 − = lim →0 9 − ( +−) + √9 −
= lim
→0
−1
9 − ( + ) + √9 − =
−1
2√9 −
Domain of = (−∞ 9], domain of 0 = (−∞ 9).
28. 0() = lim
→0
( + ) − ()
= lim →0
( + )2 − 1
2( + ) − 3 −
2 − 1
2 − 3
= lim
→0
[( + )2 − 1](2 − 3) − [2( + ) − 3](2 − 1)
[2( + ) − 3](2 − 3)
= lim
→0
(2 + 2 + 2 − 1)(2 − 3) − (2 + 2 − 3)(2 − 1)
[2( + ) − 3](2 − 3)
= lim
→0
(23 + 42 + 22 − 2 − 32 − 6 − 32 + 3) − (23 + 22 − 32 − 2 − 2 + 3)
(2 + 2 − 3)(2 − 3)
= lim
→0
42 + 22 − 6 − 32 − 22 + 2
(2 + 2 − 3)(2 − 3) = lim →0 (2(22+ 2 + 2 − − 6 3)(2 −3− 3) + 2)
= lim
→0
22 + 2 − 6 − 3 + 2
(2 + 2 − 3)(2 − 3) =
22 − 6 + 2
(2 − 3)2
Domain of = domain of 0 = (−∞ 3 2) ∪ ( 3 2 ∞).
29. 0() = lim
→0
( + ) − ()
= lim →0
1 − 2( + )
3 + ( + ) −
1 − 2
3 +
= lim
→0
[1 − 2( + )](3 + ) − [3 + ( + )](1 − 2)
[3 + ( + )](3 + )
= lim
→0
3 + − 6 − 22 − 6 − 2 − (3 − 6 + − 22 + − 2)
[3 + ( + )](3 + ) = lim →0 (3 +−6+ −)(3 + )
= lim
→0
−7
(3 + + )(3 + ) = lim →0 (3 + +−7)(3 + ) = (3 + −7)2
Domain of = domain of 0 = (−∞ −3) ∪ (−3 ∞).
30. 0() = lim
→0
( + ) − ()
= lim →0 ( + )32 − 32 = lim →0 [( + )32[(− +32)][( 32++3)32]2 + 32]
= lim
→0
( + )3 − 3
[( + )32 + 32] = lim →0 3 + 3 [(2++ 3 )3 2 2++332]− 3 = lim →0 [(3+2 + 3 )3 2 ++322]
= lim
→0
32 + 3 + 2
( + )32 + 32 =
32
232 =
32
12
Domain of = domain of 0 = [0 ∞). Strictly speaking, the domain of 0 is (0 ∞) because the limit that defines 0(0) does
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not exist (as a two-sided limit). But the right-hand derivative (in the sense of Exercise 64) does exist at 0, so in that sense one
could regard the domain of 0 to be [0 ∞).
31. 0() = lim
→0
( + ) − ()
= lim →0 ( + )4 − 4 = lim →0 4 + 43 + 622+ 43 + 4 − 4
= lim
→0
43 + 622 + 43 + 4
= lim →0 43 + 62 + 42 + 3 = 43
Domain of = domain of 0 = R.
32. (a)
(b) Note that the third graph in part (a) has small negative values for its slope, 0; but as → 6−, 0 → −∞.
See the graph in part (d).
(c) 0() = lim
→0
( + ) − ()
= lim
→0
6 − ( + ) − √6 −
6 6 − − ( ( + + ) + ) + √ √6 6 − −
= lim
→0
[6 − ( + )] − (6 − )
6 − ( + ) + √6 − = lim →0 √6 − −−+ √6 −
= lim
→0
−1
√6 − − + √6 − =
−1
2 √6 −
Domain of = (−∞ 6], domain of 0 = (−∞ 6).
(d)
33. (a) 0() = lim
→0
( + ) − ()
= lim →0 [( + )4 + 2( + )] − (4 + 2)
= lim
→0
4 + 43 + 622 + 43 + 4 + 2 + 2 − 4 − 2
= lim
→0
43 + 622 + 43 + 4 + 2
= lim →0 (43 + 62 + 4 2 + 3 + 2)
= lim
→0
(43 + 62 + 42 + 3 + 2) = 43 + 2
(b) Notice that 0() = 0 when has a horizontal tangent, 0() is
positive when the tangents have positive slope, and 0() is
negative when the tangents have negative slope.
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34. (a) 0() = lim
→0
( + ) − ()
= lim →0 [( + ) + 1( + )] − ( + 1) = lim →0
( + )2 + 1
+ −
2 + 1
= lim
→0
[( + )2 + 1] − ( + )(2 + 1)
( + ) = lim →0 (3 + 22 + 2+(+) −)(3 + + 2 + )
= lim
→0
2 + 2 −
( + ) = lim →0 (2(++)− 1) = lim →0 2(++)− 1 = 2−2 1, or 1 − 12
(b) Notice that 0() = 0 when has a horizontal tangent, 0() is
positive when the tangents have positive slope, and 0() is
negative when the tangents have negative slope. Both functions
are discontinuous at = 0.
35. (a) 0() is the rate at which the unemployment rate is changing with respect to time. Its units are percent unemployed
per year.
(b) To find 0(), we use lim
→0
( + ) − ()
≈
( + ) − ()
for small values of .
For 2003: 0(2003) ≈ (2004) − (2003)
2004 − 2003 =
55 − 60
1
= −05
For 2004: We estimate 0(2004) by using = −1 and = 1, and then average the two results to obtain a final estimate.
= −1 ⇒ 0(2004) ≈ (2003) − (2004)
2003 − 2004 =
60 − 55
−1
= −05;
= 1 ⇒ 0(2004) ≈ (2005) − (2004)
2005 − 2004 =
51 − 55
1
= −04.
So we estimate that 0(2004) ≈ 1 2[−05 + (−04)] = −045.
2003 2004 2005 2006 2007 2008 2009 2010 2011 2012
0() −050 −045 −045 −025 060 235 190 −020 −075 −080
36. (a) 0() is the rate at which the number of minimally invasive cosmetic surgery procedures performed in the United States is
changing with respect to time. Its units are thousands of surgeries per year.
(b) To find 0(), we use lim
→0
( + ) − ()
≈
( + ) − ()
for small values of .
For 2000: 0(2000) ≈ (2002) − (2000)
2002 − 2000
=
4897 − 5500
2
= −3015
For 2002: We estimate 0(2002) by using = −2 and = 2, and then average the two results to obtain a final estimate.
= −2 ⇒ 0(2002) ≈ (2000) − (2002)
2000 − 2002
=
5500 − 4897
−2
= −3015
= 2 ⇒ 0(2002) ≈ (2004) − (2002)
2004 − 2002 =
7470 − 4897
2
= 12865
So we estimate that 0(2002) ≈ 1 2[−3015 + 12865] = 4925.
2000 2002 2004 2006 2008 2010 2012
0() −3015 4925 106025 85675 60575 5345 737
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(c) (d) We could get more accurate values
for 0() by obtaining data for
more values of .
37. As in Exercise 35, we use one-sided difference quotients for the
first and last values, and average two difference quotients for all
other values.
14 21 28 35 42 49
() 41 54 64 72 78 83
0() 13 7 23 14 18 14 14 14 11 14 5 7
38. As in Exercise 35, we use one-sided difference quotients for the
first and last values, and average two difference quotients for all
other values. The units for 0() are grams per degree (g◦C).
155 177 200 224 244
() 372 310 198 97 −98
0() −282 −387 −453 −673 −975
39. (a) is the rate at which the percentage of the city’s electrical power produced by solar panels changes with respect to
time , measured in percentage points per year.
(b) 2 years after January 1, 2000 (January 1, 2002), the percentage of electrical power produced by solar panels was increasing
at a rate of 3.5 percentage points per year.
40. is the rate at which the number of people who travel by car to another state for a vacation changes with respect to the
price of gasoline. If the price of gasoline goes up, we would expect fewer people to travel, so we would expect to be
negative.
41. is not differentiable at = −4, because the graph has a corner there, and at = 0, because there is a discontinuity there.
42. is not differentiable at = −1, because there is a discontinuity there, and at = 2, because the graph has a corner there.
43. is not differentiable at = 1, because is not defined there, and at = 5, because the graph has a vertical tangent there.
44. is not differentiable at = −2 and = 3, because the graph has corners there, and at = 1, because there is a discontinuity
there.
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45. As we zoom in toward (−1 0), the curve appears more and more like a straight
line, so () = + || is differentiable at = −1. But no matter how much
we zoom in toward the origin, the curve doesn’t straighten out—we can’t
eliminate the sharp point (a cusp). So is not differentiable at = 0.
46. As we zoom in toward (0 1), the curve appears more and more like a straight
line, so is differentiable at = 0. But no matter how much we zoom in toward
(1 0) or (−1 0), the curve doesn’t straighten out—we can’t eliminate the sharp
point (a cusp). So is not differentiable at = ±1.
47. Call the curve with the positive -intercept and the other curve . Notice that has a maximum (horizontal tangent) at
= 0, but 6= 0, so cannot be the derivative of . Also notice that where is positive, is increasing. Thus, = and
= 0. Now 0(−1) is negative since 0 is below the -axis there and 00(1) is positive since is concave upward at = 1.
Therefore, 00(1) is greater than 0(−1).
48. Call the curve with the smallest positive -intercept and the other curve . Notice that where is positive in the first
quadrant, is increasing. Thus, = and = 0. Now 0(−1) is positive since 0 is above the -axis there and 00(1)
appears to be zero since has an inflection point at = 1. Therefore, 0(1) is greater than 00(−1).
49. = , = 0, = 00. We can see this because where has a horizontal tangent, = 0, and where has a horizontal tangent,
= 0. We can immediately see that can be neither nor 0, since at the points where has a horizontal tangent, neither
nor is equal to 0.
50. Where has horizontal tangents, only is 0, so 0 = . has negative tangents for 0 and is the only graph that is
negative for 0, so 0 = . has positive tangents on R (except at = 0), and the only graph that is positive on the same
domain is , so 0 = . We conclude that = , = 0, = 00, and = 000.
51. We can immediately see that is the graph of the acceleration function, since at the points where has a horizontal tangent,
neither nor is equal to 0. Next, we note that = 0 at the point where has a horizontal tangent, so must be the graph of
the velocity function, and hence, 0 = . We conclude that is the graph of the position function.
52. must be the jerk since none of the graphs are 0 at its high and low points. is 0 where has a maximum, so 0 = . is 0
where has a maximum, so 0 = . We conclude that is the position function, is the velocity, is the acceleration, and is
the jerk.
53. 0() = lim
→0
( + ) − ()
= lim →0 [3( + )2 + 2( + ) + 1] − (32 + 2 + 1)
= lim
→0
(32 + 6 + 32 + 2 + 2 + 1) − (32 + 2 + 1)
= lim →0 6 + 32 + 2
= lim
→0
(6 + 3 + 2)
= lim →0(6 + 3 + 2) = 6 + 2
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00() = lim
→0
0( + ) − 0()
= lim →0 [6( + ) + 2] − (6 + 2) = lim →0 (6 + 6 + 2) − (6 + 2)
= lim
→0
6
= lim →0 6 = 6
We see from the graph that our answers are reasonable because the graph of
0 is that of a linear function and the graph of 00 is that of a constant
function.
54. 0() = lim
→0
( + ) − ()
= lim →0 [( + )3 − 3( + )] − (3 − 3)
= lim
→0
(3 + 32 + 32 + 3 − 3 − 3) − (3 − 3)
= lim →0 32 + 32 + 3 − 3
= lim
→0
(32 + 3 + 2 − 3)
= lim →0(32 + 3 + 2 − 3) = 32 − 3
00() = lim
→0
0( + ) − 0()
= lim →0 [3( + )2 −3] − (32 − 3) = lim →0 (32 + 6 + 32− 3) − (32 − 3)
= lim
→0
6 + 32
= lim →0 (6+ 3) = lim →0(6 + 3) = 6
We see from the graph that our answers are reasonable because the graph of
0 is that of an even function ( is an odd function) and the graph of 00 is
that of an odd function. Furthermore, 0 = 0 when has a horizontal
tangent and 00 = 0 when 0 has a horizontal tangent.
55. 0() = lim
→0
( + ) − ()
= lim →0
2( + )2 − ( + )3 − (22 − 3)
= lim
→0
(4 + 2 − 32 − 3 − 2)
= lim →0 (4 + 2 − 32 − 3 − 2) = 4 − 32
00() = lim
→0
0( + ) − 0()
= lim →0
4( + ) − 3( + )2 − (4 − 32)
= lim →0 (4 − 6 − 3)
= lim
→0
(4 − 6 − 3) = 4 − 6
000() = lim
→0
00( + ) − 00()
= lim →0 [4 − 6( + )] − (4 − 6) = lim →0 −6 = lim →0 (−6) = −6
(4)() = lim
→0
000( + ) − 000()
= lim →0 −6 −(−6) = lim →0 0 = lim →0 (0) = 0
The graphs are consistent with the geometric interpretations of the
derivatives because 0 has zeros where has a local minimum and a local
maximum, 00 has a zero where 0 has a local maximum, and 000 is a
constant function equal to the slope of 00.
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56. (a) Since we estimate the velocity to be a maximum
at = 10, the acceleration is 0 at = 10.
(b) Drawing a tangent line at = 10 on the graph of , appears to decrease by 10 fts2 over a period of 20 s.
So at = 10 s, the jerk is approximately −1020 = −05 (fts2)s or fts3.
57. (a) Note that we have factored − as the difference of two cubes in the third step.
0() = lim
→
() − ()
−
= lim
→
13 − 13
−
= lim
→
13 − 13
(13 − 13)(23 + 1313 + 23)
= lim
→
1
23 + 1313 + 23 =
1
323 or 1 3−23
(b) 0(0) = lim
→0
(0 + ) − (0)
= lim →0
√3 − 0
= lim →0 213 . This function increases without bound, so the limit does not
exist, and therefore 0(0) does not exist.
(c) lim
→0
| 0()| = lim
→0
1
323 = ∞ and is continuous at = 0 (root function), so has a vertical tangent at = 0.
58. (a) 0(0) = lim
→0
() − (0)
− 0
= lim
→0
23 − 0
= lim
→0
1
13 , which does not exist.
(b) 0() = lim
→
() − ()
−
= lim
→
23 − 23
−
= lim
→
(13 − 13)(13 + 13)
(13 − 13)(23 + 1313 + 23)
= lim
→
13 + 13
23 + 1313 + 23 =
213
323 =
2
313 or 2 3−13
(c) () = 23 is continuous at = 0 and
lim
→0
|0()| = lim
→0
2
3 ||13 = ∞. This shows that
has a vertical tangent line at = 0.
(d)
59. () = | − 6| = − (−6− 6) if if − − 6 6 ≥ 6 0 = 6 −−6 if if ≥ 6 6
So the right-hand limit is lim
→6+
() − (6)
− 6
= lim
→6+
| − 6| − 0
− 6
= lim
→6+
− 6
− 6
= lim
→6+
1 = 1, and the left-hand limit
is lim
→6−
() − (6)
− 6
= lim
→6−
| − 6| − 0
− 6
= lim
→6−
6 −
− 6
= lim
→6−
(−1) = −1. Since these limits are not equal,
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0(6) = lim
→6
() − (6)
− 6
does not exist and is not differentiable at 6.
However, a formula for 0 is 0() = 1 −1 if if 66
Another way of writing the formula is 0() = − 6
| − 6|.
60. () = [[]] is not continuous at any integer , so is not differentiable
at by the contrapositive of Theorem 4. If is not an integer, then
is constant on an open interval containing , so 0() = 0. Thus,
0() = 0, not an integer.
61. (a) () = || = −22 if if ≥ 0 0 (b) Since () = 2 for ≥ 0, we have 0() = 2 for 0.
[See Exercise 19(d).] Similarly, since () = −2 for 0,
we have 0() = −2 for 0. At = 0, we have
0(0) = lim
→0
() − (0)
− 0
= lim
→0
||
= lim
→0
|| = 0
So is differentiable at 0. Thus, is differentiable for all .
(c) From part (b), we have 0() = −22 if if ≥ 0 0 = 2 ||.
62. (a) || = − if if ≥ 0 0
so () = + || = 20 if if ≥ 0 0 .
Graph the line = 2 for ≥ 0 and graph = 0 (the x-axis) for 0.
(b) is not differentiable at = 0 because the graph has a corner there, but
is differentiable at all other values; that is, is differentiable on (−∞ 0) ∪ (0 ∞).
(c) () = 2 0 if if ≥ 0 0 ⇒ 0() = 20 if if 0 0
Another way of writing the formula is 0() = 1 + sgn for 6= 0.
63. (a) If is even, then
0(−) = lim
→0
(− + ) − (−)
= lim →0 [−( − )] − (−)
= lim
→0
( − ) − ()
= − lim →0 ( − −)− () [let ∆ = −]
= − lim
∆→0
( + ∆) − ()
∆ = − 0()
Therefore, 0 is odd.
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(b) If is odd, then
0(−) = lim
→0
(− + ) − (−)
= lim →0 [−( − )] − (−)
= lim
→0
−( − ) + ()
= lim →0 ( − −)− () [let ∆ = −]
= lim
∆→0
( + ∆) − ()
∆ = 0()
Therefore, 0 is even.
64. (a) −0 (4) = lim
→0−
(4 + ) − (4)
= lim →0− 5 − (4 + ) − 1
= lim
→0−
−
= −1
and
+0 (4) = lim
→0+
(4 + ) − (4)
= lim →0+
1
5 − (4 + ) − 1
= lim
→0+
1 − (1 − )
(1 − ) = lim →0+ 1 −1 = 1
(b)
(c) () =
0 if ≤ 0
5 − if 0 4
1(5 − ) if ≥ 4
At 4 we have lim
→4−
() = lim
→4−
(5 − ) = 1 and lim
→4+
() = lim
→4+
1
5 −
= 1, so lim
→4
() = 1 = (4) and is
continuous at 4. Since (5) is not defined, is discontinuous at 5. These expressions show that is continuous on the
intervals (−∞ 0), (0 4), (4 5) and (5 ∞). Since lim
→0+
() = lim
→0+
(5 − ) = 5 6= 0 = lim
→0−
(), lim
→0
() does
not exist, so is discontinuous (and therefore not differentiable) at 0.
(d) From (a), is not differentiable at 4 since −0 (4) 6= +0 (4), and from (c), is not differentiable at 0 or 5.
65. These graphs are idealizations conveying the spirit of the problem. In reality, changes in speed are not instantaneous, so the
graph in (a) would not have corners and the graph in (b) would be continuous.
(a) (b)
66. (a)
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(b) The initial temperature of the water is close to room temperature because of
the water that was in the pipes. When the water from the hot water tank
starts coming out, is large and positive as increases to the
temperature of the water in the tank. In the next phase, = 0 as the
water comes out at a constant, high temperature. After some time,
becomes small and negative as the contents of the hot water tank are
exhausted. Finally, when the hot water has run out, is once again 0 as
the water maintains its (cold) temperature.
(c)
67. In the right triangle in the diagram, let ∆ be the side opposite angle and ∆
the side adjacent to angle . Then the slope of the tangent line
is = ∆∆ = tan . Note that 0 2 . We know (see Exercise 19)
that the derivative of () = 2 is 0() = 2. So the slope of the tangent to
the curve at the point (1 1) is 2. Thus, is the angle between 0 and 2 whose
tangent is 2; that is, = tan−1 2 ≈ 63◦.
2 Review
1. False. Limit Law 2 applies only if the individual limits exist (these don’t).
2. False. Limit Law 5 cannot be applied if the limit of the denominator is 0 (it is).
3. True. Limit Law 5 applies.
4. False. 2 − 9
− 3
is not defined when = 3, but + 3 is.
5. True. lim
→3
2 − 9
− 3
= lim
→3
( + 3)( − 3)
( − 3) = lim →3( + 3)
6. True. The limit doesn’t exist since ()() doesn’t approach any real number as approaches 5.
(The denominator approaches 0 and the numerator doesn’t.)
7. False. Consider lim
→5
( − 5)
− 5
or lim
→5
sin( − 5)
− 5
. The first limit exists and is equal to 5. By Example 2.2.3, we know that
the latter limit exists (and it is equal to 1).
8. False. If () = 1, () = −1, and = 0, then lim
→0
() does not exist, lim
→0
() does not exist, but
lim
→0
[() + ()] = lim
→0
0 = 0 exists.
9. True. Suppose that lim
→
[() + ()] exists. Now lim
→
() exists and lim
→
() does not exist, but
lim
→
() = lim
→
{[() + ()] − ()} = lim
→
[() + ()] − lim
→
() [by Limit Law 2], which exists, and
we have a contradiction. Thus, lim
→
[() + ()] does not exist.
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10. False. Consider lim
→6
[()()] = lim
→6 ( − 6) −1 6. It exists (its value is 1) but (6) = 0 and (6) does not exist,
so (6)(6) 6= 1.
11. True. A polynomial is continuous everywhere, so lim
→
() exists and is equal to ().
12. False. Consider lim
→0
[() − ()] = lim
→0 12 − 14 . This limit is −∞ (not 0), but each of the individual functions
approaches ∞.
13. True. See Figure 2.6.8.
14. False. Consider () = sin for ≥ 0. lim
→∞
() 6= ±∞ and has no horizontal asymptote.
15. False. Consider () = 1 2( − 1) if if 6= 1 = 1
16. False. The function must be continuous in order to use the Intermediate Value Theorem. For example, let
() = 1 −1 if if 0 ≤= 3 3 There is no number ∈ [0 3] with () = 0.
17. True. Use Theorem 2.5.8 with = 2, = 5, and () = 42 − 11. Note that (4) = 3 is not needed.
18. True. Use the Intermediate Value Theorem with = −1, = 1, and = , since 3 4.
19. True, by the definition of a limit with = 1.
20. False. For example, let () = 22 + 1 if if 6= 0 = 0
Then () 1 for all , but lim
→0
() = lim
→0
2 + 1 = 1.
21. False. See the note after Theorem 2.8.4.
22. True. 0() exists ⇒ is differentiable at ⇒ is continuous at ⇒ lim
→
() = ().
23. False. 2
2 is the second derivative while 2 is the first derivative squared. For example, if = ,
then 2
2 = 0, but 2 = 1.
24. True. () = 10 − 102 + 5 is continuous on the interval [0 2], (0) = 5, (1) = −4, and (2) = 989. Since
−4 0 5, there is a number in (0 1) such that () = 0 by the Intermediate Value Theorem. Thus, there is a
root of the equation 10 − 102 + 5 = 0 in the interval (0 1). Similarly, there is a root in (1 2).
25. True. See Exercise 2.5.72(b).
26. False See Exercise 2.5.72(b).
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1. (a) (i) lim
→2+
() = 3 (ii) lim
→−3+
() = 0
(iii) lim
→−3
() does not exist since the left and right limits are not equal. (The left limit is −2.)
(iv) lim
→4
() = 2
(v) lim
→0
() = ∞ (vi) lim
→2−
() = −∞
(vii) lim
→∞
() = 4 (viii) lim
→−∞
() = −1
(b) The equations of the horizontal asymptotes are = −1 and = 4.
(c) The equations of the vertical asymptotes are = 0 and = 2.
(d) is discontinuous at = −3, 0, 2, and 4. The discontinuities are jump, infinite, infinite, and removable, respectively.
2. lim
→−∞
() = −2, lim
→∞
() = 0, lim
→−3
() = ∞,
lim
→3−
() = −∞, lim
→3+
() = 2,
is continuous from the right at 3
3. Since the exponential function is continuous, lim
→1
3− = 1−1 = 0 = 1.
4. Since rational functions are continuous, lim
→3
2 − 9
2 + 2 − 3 =
32 − 9
32 + 2(3) − 3 =
0
12
= 0.
5. lim
→−3
2 − 9
2 + 2 − 3 = lim →−3 ( ( + 3)( + 3)( − − 3) 1) = lim →−3 − − 3 1 = −−33 −− 31 = −−64 = 32
6. lim
→1+
2 − 9
2 + 2 − 3 = −∞ since 2 + 2 − 3 → 0+ as → 1+ and 2 + 2 2 −9− 3 0 for 1 3.
7. lim
→0
( − 1)3 + 1
= lim →0
3 − 32 + 3 − 1 + 1
= lim →0 3 − 32 + 3 = lim →0 2 − 3 + 3 = 3
Another solution: Factor the numerator as a sum of two cubes and then simplify.
lim
→0
( − 1)3 + 1
= lim →0 ( − 1) 3 + 13 = lim →0 [( − 1) + 1] ( − 1) 2 − 1( − 1) + 12
= lim
→0
( − 1)2 − + 2 = 1 − 0 + 2 = 3
8. lim
→2
2 − 4
3 − 8 = lim →2 ( −(2)( + 2)( 2 + 2 −2) + 4) = lim →2 2 + 2 + 2 + 4 = 4 + 4 + 4 2 + 2 = 12 4 = 1 3
9. lim
→9
√
( − 9)4 = ∞ since ( − 9)4 → 0+ as → 9 and ( √−9)4 0 for 6= 9.
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10. lim
→4+
4 −
|4 − | = lim →4+ −4 (4−−) = lim →4+ −11 = −1
11. lim
→1
4 − 1
3 + 52 − 6 = lim →1 ((2+ 1)( 2 + 52−−6) 1) = lim →1 (2+ 1)( ( + 6)( + 1)( −1) − 1) = lim →1 (2+ 1)( ( + 6) + 1) = 2(2) 1(7) = 47
12. lim
→3
√ + 6 −
3 − 32 = lim →3 √2(+ 6 −−3) · √√ + 6 + + 6 + = lim →3 2((√−3) + 6 ) √2+ 6 + − 2
= lim
→3
+ 6 − 2
2( − 3) √ + 6 + = lim →3 2(−−(3)2−√−+ 6 + 6) = lim →3 2(−−(3) −3)( √+ 6 + + 2)
= lim
→3
−( + 2)
2√ + 6 + = −
5
9(3 + 3) = −
5
54
13. Since is positive, √2 = || = . Thus,
lim
→∞
√2 − 9
2 − 6
= lim
→∞
√2 − 9√2
(2 − 6) = lim →∞
1 − 92
2 − 6 =
√1 − 0
2 − 0
=
1 2
14. Since is negative, √2 = || = −. Thus,
lim
→−∞
√2 − 9
2 − 6
= lim
→−∞
√2 − 9√2
(2 − 6)(−) = lim →−∞
1 − 92
−2 + 6 =
√1 − 0
−2 + 0
= −
1 2
15. Let = sin . Then as → −, sin → 0+, so → 0+. Thus, lim
→−
ln(sin ) = lim
→0+
ln = −∞.
16. lim
→−∞
1 − 22 − 4
5 + − 34 = lim →−∞ (1 (5 + − 2−2 −344)) 44 = lim →−∞ 1 5 4 4 − + 1 2 2 3 − − 1 3 = 00 + 0 − 0 −− 13 = −−13 = 13
17. lim
→∞
√2 + 4 + 1 − = lim
→∞√2 + 41 + 1 − · √√ 2 2 + 4 + 4 + 1 + + 1 + = lim →∞ (√22+ 4 + 4 + 1) + 1 + − 2
= lim
→∞
(4 + 1)
( √2 + 4 + 1 + ) divide by = √2 for 0
= lim
→∞
4 + 1
1 + 4 + 12 + 1 =
4 + 0
√1 + 0 + 0 + 1 =
4 2
= 2
18. Let = − 2 = (1 − ). Then as → ∞, → −∞, and lim
→∞
−2 = lim
→−∞
= 0.
19. Let = 1. Then as → 0+, → ∞, and lim
→0+
tan−1(1) = lim
→∞
tan−1 =
2
.
20. lim
→1 −1 1 + 2 − 31 + 2 = lim →1 −1 1 + ( − 1)( 1 − 2) = lim →1 ( −1)( −2− 2) + ( − 1)( 1 − 2)
= lim
→1 ( −1)( −1− 2) = lim →1 −1 2 = 1 −1 2 = −1
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© Cengage Learning. All Rights Reserved.CHAPTER 2 REVIEW ¤ 155
21. From the graph of = cos2 2, it appears that = 0 is the horizontal
asymptote and = 0 is the vertical asymptote. Now 0 ≤ (cos )2 ≤ 1 ⇒
0 2
≤ cos2
2 ≤ 12 ⇒ 0 ≤ cos 22 ≤ 12 . But →lim±∞0 = 0 and
lim
→±∞
1 2
= 0, so by the Squeeze Theorem, lim
→±∞
cos2
2 = 0.
Thus, = 0 is the horizontal asymptote. lim
→0
cos2
2 = ∞ because cos2 → 1 and 2 → 0+ as → 0, so = 0 is the
vertical asymptote.
22. From the graph of = () = √2 + + 1 − √2 − , it appears that there are 2 horizontal asymptotes and possibly 2
vertical asymptotes. To obtain a different form for , let’s multiply and divide it by its conjugate.
1() = √2 + + 1 − √2 −
√2 + + 1 + √2 −
√2 + + 1 + √2 − =
(2 + + 1) − (2 − )
√2 + + 1 + √2 −
=
2 + 1
√2 + + 1 + √2 −
Now
lim
→∞
1() = lim
→∞
2 + 1
√2 + + 1 + √2 −
= lim
→∞
2 + (1)
1 + (1) + (12) + 1 − (1) [since √2 = for 0]
=
2
1 + 1
= 1,
so = 1 is a horizontal asymptote. For 0, we have √2 = || = −, so when we divide the denominator by ,
with 0, we get
√2 + + 1 + √2 −
= −
√2 + + 1 + √2 −
√2 = − 1 + 1 + 12 + 1 − 1
Therefore,
lim
→−∞
1() = lim
→−∞
2 + 1
√2 + + 1 + √2 − = lim →∞ − 1 + (1) + (1 2 + (1 2) + ) 1 − (1)
=
2
−(1 + 1) = −1
so = −1 is a horizontal asymptote.
The domain of is (−∞ 0] ∪ [1 ∞). As → 0−, () → 1, so
= 0 is not a vertical asymptote. As → 1+, () → √3, so = 1
is not a vertical asymptote and hence there are no vertical asymptotes.
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23. Since 2 − 1 ≤ () ≤ 2 for 0 3 and lim
→1
(2 − 1) = 1 = lim
→1
2, we have lim
→1
() = 1 by the Squeeze Theorem.
24. Let () = −2, () = 2 cos12 and () = 2. Then since cos12 ≤ 1 for 6= 0, we have
() ≤ () ≤ () for 6= 0, and so lim
→0
() = lim
→0
() = 0 ⇒ lim
→0
() = 0 by the Squeeze Theorem.
25. Given 0, we need 0 such that if 0 | − 2| , then |(14 − 5) − 4| . But |(14 − 5) − 4| ⇔
|−5 + 10| ⇔ |−5| | − 2| ⇔ | − 2| 5. So if we choose = 5, then 0 | − 2| ⇒
|(14 − 5) − 4| . Thus, lim
→2
(14 − 5) = 4 by the definition of a limit.
26. Given 0 we must find 0 so that if 0 | − 0| , then |√3 − 0| . Now |√3 − 0| = |√3 | ⇒
|| = |√3 |3 3. So take = 3. Then 0 | − 0| = || 3 ⇒ |√3 − 0| = |√3 | = 3 || √3 3 = .
Therefore, by the definition of a limit, lim
→0
√3 = 0.
27. Given 0, we need 0 so that if 0 | − 2| , then 2 − 3 − (−2) . First, note that if | − 2| 1, then
−1 − 2 1, so 0 − 1 2 ⇒ | − 1| 2. Now let = min {2 1}. Then 0 | − 2| ⇒
2 − 3 − (−2) = |( − 2)( − 1)| = | − 2| | − 1| (2)(2) = .
Thus, lim
→2
(2 − 3) = −2 by the definition of a limit.
28. Given 0, we need 0 such that if 0 − 4 , then 2√ − 4 . This is true ⇔ √ − 4 2 ⇔
− 4 4 2. So if we choose = 4 2, then 0 − 4 ⇒ 2√ − 4 . So by the definition of a limit,
lim
→4+
2√ − 4 = ∞.
29. (a) () = √− if 0, () = 3 − if 0 ≤ 3, () = ( − 3)2 if 3.
(i) lim
→0+
() = lim
→0+
(3 − ) = 3 (ii) lim
→0−
() = lim
→0−
√− = 0
(iii) Because of (i) and (ii), lim
→0
() does not exist. (iv) lim
→3−
() = lim
→3−
(3 − ) = 0
(v) lim
→3+
() = lim
→3+
( − 3)2 = 0 (vi) Because of (iv) and (v), lim
→3
() = 0.
(b) is discontinuous at 0 since lim
→0
() does not exist.
is discontinuous at 3 since (3) does not exist.
(c)
30. (a) () = 2 − 2 if 0 ≤ ≤ 2, () = 2 − if 2 ≤ 3, () = − 4 if 3 4, () = if ≥ 4.
Therefore, lim
→2−
() = lim
→2−
2 − 2 = 0 and lim
→2+
() = lim
→2+
(2 − ) = 0. Thus, lim
→2
() = 0 = (2),
so is continuous at 2. lim
→3−
() = lim
→3−
(2 − ) = −1 and lim
→3+
() = lim
→3+
( − 4) = −1. Thus,
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lim
→3
() = −1 = (3), so is continuous at 3.
lim
→4−
() = lim
→4−
( − 4) = 0 and lim
→4+
() = lim
→4+
= .
Thus, lim
→4
() does not exist, so is discontinuous at 4. But
lim
→4+
() = = (4), so is continuous from the right at 4.
(b)
31. sin and are continuous on R by Theorem 2.5.7. Since is continuous on R, sin is continuous on R by Theorem 2.5.9.
Lastly, is continuous on R since it’s a polynomial and the product sin is continuous on its domain R by Theorem 2.5.4.
32. 2 − 9 is continuous on R since it is a polynomial and √ is continuous on [0 ∞) by Theorem 2.5.7, so the composition
√2 − 9 is continuous on | 2 − 9 ≥ 0 = (−∞ −3] ∪ [3 ∞) by Theorem 2.5.9. Note that 2 − 2 6= 0 on this set and
so the quotient function () =
√2 − 9
2 − 2 is continuous on its domain, (−∞ −3] ∪ [3 ∞) by Theorem 2.5.4.
33. () = 5 − 3 + 3 − 5 is continuous on the interval [1 2], (1) = −2, and (2) = 25. Since −2 0 25, there is a
number in (1 2) such that () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation
5 − 3 + 3 − 5 = 0 in the interval (1 2).
34. () = cos √ − + 2 is continuous on the interval [0 1], (0) = 2, and (1) ≈ −02. Since −02 0 2, there is a
number in (0 1) such that () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation
cos √ − + 2 = 0, or cos √ = − 2, in the interval (0 1).
35. (a) The slope of the tangent line at (2 1) is
lim
→2
() − (2)
− 2
= lim
→2
9 − 22 − 1
− 2
= lim
→2
8 − 22
− 2
= lim
→2
−2(2 − 4)
− 2
= lim
→2
−2( − 2)( + 2)
− 2
= lim
→2
[−2( + 2)] = −2 · 4 = −8
(b) An equation of this tangent line is − 1 = −8( − 2) or = −8 + 17.
36. For a general point with -coordinate , we have
= lim
→
2(1 − 3) − 2(1 − 3)
−
= lim
→
2(1 − 3) − 2(1 − 3)
(1 − 3)(1 − 3)( − ) = lim → (1 − 3)(1 6(−−3))( − )
= lim
→
6
(1 − 3)(1 − 3) =
6
(1 − 3)2
For = 0, = 6 and (0) = 2, so an equation of the tangent line is − 2 = 6( − 0) or = 6 + 2 For = −1, = 3 8
and (−1) = 1 2, so an equation of the tangent line is − 1 2 = 3 8( + 1) or = 3 8 + 7 8.
37. (a) = () = 1 + 2 + 24. The average velocity over the time interval [1 1 + ] is
ave =
(1 + ) − (1)
(1 + ) − 1 =
1 + 2(1 + ) + (1 + )24 − 134
=
10 + 2
4 =
10 +
4
[continued]
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So for the following intervals the average velocities are:
(i) [1 3]: = 2, ave = (10 + 2)4 = 3 ms (ii) [1 2]: = 1, ave = (10 + 1)4 = 275 ms
(iii) [1 15]: = 05, ave = (10 + 05)4 = 2625 ms (iv) [1 11]: = 01, ave = (10 + 01)4 = 2525 ms
(b) When = 1, the instantaneous velocity is lim
→0
(1 + ) − (1)
= lim →0 10 + 4 = 10 4 = 25 ms.
38. (a) When increases from 200 in3 to 250 in3, we have ∆ = 250 − 200 = 50 in3, and since = 800 ,
∆ = (250) − (200) = 800
250
−
800
200
= 32 − 4 = −08 lbin2. So the average rate of change
is ∆
∆ =
−08
50
= −0016 lbin2
in3 .
(b) Since = 800 , the instantaneous rate of change of with respect to is
lim
→0
∆
∆ = lim →0 ( + ) − () = lim →0 800( + ) − 800 = lim →0 800 [ (−+()+ )]
= lim
→0
−800
( + ) = −
800
2
which is inversely proportional to the square of .
39. (a) 0(2) = lim
→2
() − (2)
− 2
= lim
→2
3 − 2 − 4
− 2
= lim
→2
( − 2)(2 + 2 + 2)
− 2
= lim
→2
(2 + 2 + 2) = 10
(c)
(b) − 4 = 10( − 2) or = 10 − 16
40. 26 = 64, so () = 6 and = 2.
41. (a) 0() is the rate at which the total cost changes with respect to the interest rate. Its units are dollars(percent per year).
(b) The total cost of paying off the loan is increasing by $1200(percent per year) as the interest rate reaches 10%. So if the
interest rate goes up from 10% to 11%, the cost goes up approximately $1200.
(c) As increases, increases. So 0() will always be positive.
42. 43. 44.
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45. (a) 0() = lim
→0
( + ) − ()
= lim →0
3 − 5( + ) − √3 − 5
3 − 5( + ) + √3 − 5
3 − 5( + ) + √3 − 5
= lim
→0
[3 − 5( + )] − (3 − 5)
3 − 5( + ) + √3 − 5 = lim →0 3 − 5( +−5) + √3 − 5 = 2√3−−5 5
(b) Domain of : (the radicand must be nonnegative) 3 − 5 ≥ 0 ⇒
5 ≤ 3 ⇒ ∈ −∞ 3 5
Domain of 0: exclude 3 5 because it makes the denominator zero;
∈ −∞ 3 5
(c) Our answer to part (a) is reasonable because 0() is always negative and
is always decreasing.
46. (a) As → ±∞, () = (4 − )(3 + ) → −1, so there is a horizontal
asymptote at = −1. As → −3+, () → ∞, and as → −3−,
() → −∞. Thus, there is a vertical asymptote at = −3.
(b) Note that is decreasing on (−∞ −3) and (−3 ∞), so 0 is negative on
those intervals. As → ±∞, 0 → 0. As → −3− and as → −3+,
0 → −∞.
(c) 0() = lim
→0
( + ) − ()
= lim →0
4 − ( + )
3 + ( + ) −
4 −
3 +
= lim →0 (3 + ) [4 −([3 + ( + )]+−(4 )] (3 + − ) [3 + ( ) + )]
= lim
→0
(12 − 3 − 3 + 4 − 2 − ) − (12 + 4 + 4 − 3 − 2 − )
[3 + ( + )](3 + )
= lim
→0
−7
[3 + ( + )] (3 + ) = lim →0 [3 + ( +−7)] (3 + ) = −(3 +7)2
(d) The graphing device confirms our graph in part (b).
47. is not differentiable: at = −4 because is not continuous, at = −1 because has a corner, at = 2 because is not
continuous, and at = 5 because has a vertical tangent.
48. The graph of has tangent lines with positive slope for 0 and negative slope for 0, and the values of fit this pattern,
so must be the graph of the derivative of the function for . The graph of has horizontal tangent lines to the left and right of
the -axis and has zeros at these points. Hence, is the graph of the derivative of the function for . Therefore, is the graph
of , is the graph of 0, and is the graph of 00.
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49. Domain: (−∞ 0) ∪ (0 ∞); lim
→0−
() = 1; lim
→0+
() = 0;
0() 0 for all in the domain; lim
→−∞
0() = 0; lim
→∞
0() = 1
50. (a) 0() is the rate at which the percentage of Americans under the age of 18 is changing with respect to time. Its units are
percent per year (%yr).
(b) To find 0(), we use lim
→0
( + ) − ()
≈
( + ) − ()
for small values of .
For 1950: 0(1950) ≈ (1960) − (1950)
1960 − 1950
=
357 − 311
10
= 046
For 1960: We estimate 0(1960) by using = −10 and = 10, and then average the two results to obtain a
final estimate.
= −10 ⇒ 0(1960) ≈ (1950) − (1960)
1950 − 1960
=
311 − 357
−10
= 046
= 10 ⇒ 0(1960) ≈ (1970) − (1960)
1970 − 1960 =
340 − 357
10
= −017
So we estimate that 0(1960) ≈ 1 2[046 + (−017)] = 0145.
1950 1960 1970 1980 1990 2000 2010
0() 0460 0145 −0385 −0415 −0115 −0085 −0170
(c)
(d) We could get more accurate values for 0() by obtaining data for the mid-decade years 1955, 1965, 1975, 1985, 1995, and
2005.
51. 0() is the rate at which the number of US $20 bills in circulation is changing with respect to time. Its units are billions of
bills per year. We use a symmetric difference quotient to estimate 0(2000).
0(2000) ≈ (2005) − (1995)
2005 − 1995
=
577 − 421
10
= 0156 billions of bills per year (or 156 million bills per year).
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© Cengage Learning. All Rights Reserved.CHAPTER 2 REVIEW ¤ 161
52. (a) Drawing slope triangles, we obtain the following estimates: 0(1950) ≈ 110 1 = 011, 0(1965) ≈ −10 16 = −016,
and 0(1987) ≈ 010 2 = 002.
(b) The rate of change of the average number of children born to each woman was increasing by 011 in 1950, decreasing
by 016 in 1965, and increasing by 002 in 1987.
(c) There are many possible reasons:
• In the baby-boom era (post-WWII), there was optimism about the economy and family size was rising.
• In the baby-bust era, there was less economic optimism, and it was considered less socially responsible to have a
large family.
• In the baby-boomlet era, there was increased economic optimism and a return to more conservative attitudes.
53. |()| ≤ () ⇔ −() ≤ () ≤ () and lim
→
() = 0 = lim
→
−().
Thus, by the Squeeze Theorem, lim
→
() = 0.
54. (a) Note that is an even function since () = (−). Now for any integer ,
[[]] + [[−]] = − = 0, and for any real number which is not an integer,
[[]] + [[−]] = [[]] + (− [[]] − 1) = −1. So lim
→
() exists (and is equal to −1)
for all values of .
(b) is discontinuous at all integers.
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© Cengage Learning. All Rights Reserved.PROBLEMS PLUS
1. Let = √6 , so = 6. Then → 1 as → 1, so
lim
→1
√3 − 1
√ − 1 = lim →1 2 3 − − 1 1 = lim →1 ( −(1) ( − 1)( 2 ++ 1) + 1) = lim →1 2 ++ 1 + 1 = 12 1 + 1 + 1 + 1 = 23.
Another method: Multiply both the numerator and the denominator by (√ + 1)√3 2 + √3 + 1.
2. First rationalize the numerator: lim
→0
√ + − 2
·
√ + + 2
√ + + 2 = lim →0 √++ − + 2 4 . Now since the denominator
approaches 0 as → 0, the limit will exist only if the numerator also approaches 0 as → 0. So we require that
(0) + − 4 = 0 ⇒ = 4. So the equation becomes lim
→0
√ + 4 + 2 = 1 ⇒ √4 + 2 = 1 ⇒ = 4.
Therefore, = = 4.
3. For − 1
2 1 2, we have 2 − 1 0 and 2 + 1 0, so |2 − 1| = −(2 − 1) and |2 + 1| = 2 + 1.
Therefore, lim
→0
|2 − 1| − |2 + 1|
= lim
→0
−(2 − 1) − (2 + 1)
= lim
→0
−4
= lim
→0
(−4) = −4.
4. Let be the midpoint of , so the coordinates of are 1 2 1 22 since the coordinates of are 2. Let = (0 ).
Since the slope = 2
= , = −
1
(negative reciprocal). But =
12
2 −
12
− 0
=
2 − 2
, so we conclude that
−1 = 2 − 2 ⇒ 2 = 2 + 1 ⇒ = 1
22 + 1 2. As → 0, → 1 2 and the limiting position of is 0 1 2.
5. (a) For 0 1, [[]] = 0, so [[]]
= 0, and lim
→0+
[[]]
= 0. For −1 0, [[]] = −1, so [[]]
=
−1
, and
lim
→0−
[[]]
= lim
→0− −1 = ∞. Since the one-sided limits are not equal, lim →0 [[ ]] does not exist.
(b) For 0, 1 − 1 ≤ [[1]] ≤ 1 ⇒ (1 − 1) ≤ [[1]] ≤ (1) ⇒ 1 − ≤ [[1]] ≤ 1.
As → 0+, 1 − → 1, so by the Squeeze Theorem, lim
→0+
[[1]] = 1.
For 0, 1 − 1 ≤ [[1]] ≤ 1 ⇒ (1 − 1) ≥ [[1]] ≥ (1) ⇒ 1 − ≥ [[1]] ≥ 1.
As → 0−, 1 − → 1, so by the Squeeze Theorem, lim
→0−
[[1]] = 1.
Since the one-sided limits are equal, lim
→0
[[1]] = 1.
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6. (a) [[]]2 + [[]]2 = 1. Since [[]]2 and [[]]2 are positive integers or 0, there are
only 4 cases:
Case (i): [[]] = 1, [[]] = 0 ⇒1 ≤ 2 and 0 ≤ 1
Case (ii): [[]] = −1, [[]] = 0 ⇒−1 ≤ 0 and 0 ≤ 1
Case (iii):[[]] = 0, [[]] = 1 ⇒0 ≤ 1 and 1 ≤ 2
Case (iv):[[]] = 0, [[]] = −1 ⇒0 ≤ 1 and −1 ≤ 0
(b) [[]]2 − [[]]2 = 3. The only integral solution of 2 − 2 = 3 is = ±2
and = ±1. So the graph is
{( ) | [[]] = ±2, [[]] = ±1} = ( ) 21 ≤≤ ≤ 23 or or −−12 ≤≤ 01 .
(c) [[ + ]]2 = 1 ⇒ [[ + ]] = ±1 ⇒ 1 ≤ + 2
or −1 ≤ + 0
(d) For ≤ + 1, [[]] = . Then [[]] + [[]] = 1 ⇒ [[]] = 1 − ⇒
1 − ≤ 2 − . Choosing integer values for produces the graph.
7. is continuous on (−∞ ) and ( ∞). To make continuous on R, we must have continuity at . Thus,
lim
→+
() = lim
→−
() ⇒ lim
→+
2 = lim
→−
( + 1) ⇒ 2 = + 1 ⇒ 2 − − 1 = 0 ⇒
[by the quadratic formula] = 1 ± √5 2 ≈ 1618 or −0618.
8. (a) Here are a few possibilities:
(b) The “obstacle” is the line = (see diagram). Any intersection of the graph of with the line = constitutes a fixed
point, and if the graph of the function does not cross the line somewhere in (0 1), then it must either start at (0 0)
(in which case 0 is a fixed point) or finish at (1 1) (in which case 1 is a fixed point).
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(c) Consider the function () = () − , where is any continuous function with domain [0 1] and range in [0 1]. We
shall prove that has a fixed point. Now if (0) = 0 then we are done: has a fixed point (the number 0), which is what
we are trying to prove. So assume (0) 6= 0. For the same reason we can assume that (1) 6= 1. Then (0) = (0) 0
and (1) = (1) − 1 0. So by the Intermediate Value Theorem, there exists some number in the interval (0 1) such
that () = () − = 0. So () = , and therefore has a fixed point.
9.
lim
→
[() + ()] = 2
lim
→
[() − ()] = 1
⇒
lim
→
() + lim
→
() = 2 (1)
lim
→
() − lim
→
() = 1 (2)
Adding equations (1) and (2) gives us 2 lim
→
() = 3 ⇒ lim
→
() = 3 2. From equation (1), lim
→
() = 1 2. Thus,
lim
→
[() ()] = lim
→
() · lim
→
() = 3 2 · 1 2 = 3 4.
10. (a) Solution 1: We introduce a coordinate system and drop a perpendicular
from , as shown. We see from ∠ that tan 2 =
1 −
, and from
∠ that tan = . Using the double-angle formula for tangents,
we get
1 −
= tan 2 = 2 tan
1 − tan2 =
2( )
1 − ( )2 . After a bit of
simplification, this becomes 1
1 −
=
2
2 − 2 ⇔ 2 = (3 − 2).
As the altitude decreases in length, the point will approach the -axis, that is, → 0, so the limiting location of
must be one of the roots of the equation (3 − 2) = 0. Obviously it is not = 0 (the point can never be to the left of
the altitude , which it would have to be in order to approach 0) so it must be 3 − 2 = 0, that is, = 2 3.
Solution 2: We add a few lines to the original diagram, as shown. Now note
that ∠ = ∠ (alternate angles; k by symmetry) and
similarly ∠ = ∠. So ∆ and ∆ are isosceles, and
the line segments , and are all of equal length. As || → 0,
and approach points on the base, and the point is seen to approach a
position two-thirds of the way between and , as above.
(b) The equation 2 = (3 − 2) calculated in part (a) is the equation of
the curve traced out by . Now as || → ∞, 2 → 2 , → 4 ,
→ 1, and since tan = , → 1. Thus, only traces out the
part of the curve with 0 ≤ 1.
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11. (a) Consider () = ( + 180◦) − (). Fix any number . If () = 0, we are done: Temperature at = Temperature
at + 180◦. If () 0, then ( + 180◦) = ( + 360◦) − ( + 180◦) = () − ( + 180◦) = −() 0.
Also, is continuous since temperature varies continuously. So, by the Intermediate Value Theorem, has a zero on the
interval [ + 180◦]. If () 0, then a similar argument applies.
(b) Yes. The same argument applies.
(c) The same argument applies for quantities that vary continuously, such as barometric pressure. But one could argue that
altitude above sea level is sometimes discontinuous, so the result might not always hold for that quantity.
12. 0() = lim
→0
( + ) − ()
= lim →0 ( + )( + ) − () = lim →0 ( + ) − () + (+ )
= lim
→0
( + ) − ()
+ lim →0 ( + ) = 0() + ()
because is differentiable and therefore continuous.
13. (a) Put = 0 and = 0 in the equation: (0 + 0) = (0) + (0) + 02 · 0 + 0 · 02 ⇒ (0) = 2(0).
Subtracting (0) from each side of this equation gives (0) = 0.
(b) 0(0) = lim
→0
(0 + ) − (0)
= lim →0
(0) + () + 02 + 02 − (0)
= lim →0 () = lim →0 () = 1
(c) 0() = lim
→0
( + ) − ()
= lim →0
() + () + 2 + 2 − ()
= lim →0 () + 2 + 2
= lim
→0 () + 2 + = 1 + 2
14. We are given that |()| ≤ 2 for all . In particular, |(0)| ≤ 0, but || ≥ 0 for all . The only conclusion is
that (0) = 0. Now
() − (0)
− 0
=
()
=
|()|
|| ≤ |2| = |2| = || ⇒ −|| ≤ () −− 0(0) ≤ ||.
But lim
→0
(−||) = 0 = lim
→0
||, so by the Squeeze Theorem, lim
→0
() − (0)
− 0
= 0. So by the definition of a derivative,
is differentiable at 0 and, furthermore, 0(0) = 0.
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