C H A P T E R 1
Limits and Their Properties
Section 1.1
A Preview of Calculus . . . . . . . . . . . . . . . . . . . 305
Section 1.2
Finding Limits Graphically and Numerically . . . . . . . 305
Section 1.3
Evaluating Limits Analytically . . . . . . . . . . . . . . . 309
Section 1.4
Continuity and One-Sided Limits
Section 1.5
Infinite Limits
Review Exercises
. . . . . . . . . . . . . 315
. . . . . . . . . . . . . . . . . . . . . . . 320
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327
C H A P T E R 1
Limits and Their Properties
Section 1.1
A Preview of Calculus
Solutions to Even-Numbered Exercises
4. Precalculus: rate of change slope 0.08
2. Calculus: velocity is not constant
Distance 20 ftsec15 seconds 300 feet
6. Precalculus: Area 2
8. Precalculus: Volume 326 54
2
2
5 5 5
10.417
2 3 4
5
5
5
5
5
5
5
1
9.145
Area 5 2
1.5 2 2.5 3 3.5 4 4.5
10. (a) Area 5 (b) You could improve the approximation by using more rectangles.
Section 1.2
2.
x
1.9
1.99
1.999
2.001
2.01
2.1
f x
0.2564
0.2506
0.2501
0.2499
0.2494
0.2439
lim
x→2
4.
x2
0.25
x2 4
3.1
3.01
3.001
2.999
2.99
2.9
f x
0.2485
0.2498
0.2500
0.2500
0.2502
0.2516
lim
1 x 2
x3
Actual limit is 14 .
0.25
x
3.9
3.99
3.999
4.001
4.01
4.1
f x
0.0408
0.0401
0.0400
0.0400
0.0399
0.0392
lim
x→4
8.
Actual limit is 14 .
x
x→3
6.
Finding Limits Graphically and Numerically
xx 1
45
0.04
x4
0.1
x
f x
lim
x→0
0.0500
0.01
0.0050
cos x 1
0.0000
x
Actual limit is 251 .
0.001
0.0005
0.001
0.01
0.0005
0.0050
0.1
0.0500
(Actual limit is 0.) (Make sure you use radian mode.)
305
306
Chapter 1
Limits and Their Properties
10. lim x2 2 3
12. lim f x lim x2 2 3
x→1
x→1
1
does not exist since the
x3
function increases and decreases
without bound as x approaches 3.
x→1
16. lim sec x 1
14. lim
x→3
18. lim sinx 0
x→0
x→1
20. Ct 0.35 0.12 t 1
(a)
1
0
5
0
(b)
t
Ct
3
3.3
3.4
3.5
3.6
3.7
4
0.59
0.71
0.71
0.71
0.71
0.71
0.71
lim Ct 0.71
t→3.5
(c)
3
2.5
2.9
3
3.1
3.5
4
0.47
0.59
0.59
0.59
0.71
0.71
0.71
t
Ct
lim Ct does not exist. The values of C jump from 0.59 to 0.71 at t 3.
t→3.5
22. You need to find such that 0 < x 2 < implies
f x 3 x2 1 3 x2 4 < 0.2. That is,
0.2
4 0.2
3.8
3.8
3.8 2
< x2 4
< x2
< x2
<
x
< x2
<
<
<
<
<
0.2
4 0.2
4.2
4.2
4.2 2
So take 4.2 2 0.0494.
Then 0 < x 2 < implies
4.2 2 < x 2 < 4.2 2
3.8 2 < x 2 < 4.2 2.
Using the first series of equivalent inequalities, you obtain
f x 3 x2 4 < 0.2.
24. lim 4 x→4
x
2
2
4
x
2 < 0.01
2
2
x
< 0.01
2
1
x 4 < 0.01
2
Hence, if 0 < x 4 < 0.02, you have
0 < x 4 < 0.02 1
x 4 < 0.01
2
2
4
x
< 0.01
2
x
2 < 0.01
2
f x L < 0.01
Section 1.2
26. lim x2 4 29
28. lim 2x 5 1
x→5
2
x→3
x 4 29 < 0.01
Finding Limits Graphically and Numerically
Given > 0:
x2 25 < 0.01
2x 5 1 < 2x 6 < 2x 3 < x 5x 5 < 0.01
0.01
x 5 < x 5
x 3
If we assume 4 < x < 6, then 0.0111 0.0009.
0.01
, you have
11
x5 <
0.01
1
<
0.01
11
x5
Hence, if 0 < x 5 < x 5x 5 < 0.01
x2 25 < 0.01
x2 4 29 < 0.01
f x L < 0.01
Hence, if 0 < x 3 < , you have
2
9 2
3
x 1
2
3x
29
3
2
3
2x 6 < 2x 5 1 < f x L < 32. lim 1 1
x→2
Given > 0:
<
<
x 1
1 1
< 0 < Hence, any > 0 will work.
Hence, for any > 0, you have
< 1 1 < f x L < < 32 Hence, let 32.
x 3 < 2
x→1
2
3x
2
Hence, let 2.
2
2
29
30. lim 3 x 9 3 1 9 3
Given > 0:
<
3
Hence, if 0 < x 1 < 2, you have
2
3x
x 1 < 32
2
3x
9 2
3
29
3
< < f x L < 34. lim x 4 2
x→4
Given > 0:
x 2 x 2 < x 4 < x→3
x 2 < x 2
x 3 0
x 3
0 < x 4 < 3 ⇒ x 4 < x 2
Given > 0:
x 2
Assuming 1 < x < 9, you can choose 3. Then,
36. lim x 3 0
⇒ x 2 < .
< < Hence, let .
Hence for 0 < x 3 < , you have
x 3 < x 3 0 < f x L < 307
308
Chapter 1
Limits and Their Properties
40. f x 38. lim x2 3x 0
x→3
Given > 0:
x2 3x 0
xx 3
x2
x3
4x 3
−3
1
2
lim f x < 4
x→3
5
−4
< The domain is all x 1, 3. The graphing utility does not
show the hole at 3, 12 .
x 3 < x
If we assume 4 < x < 2, then 4.
Hence for 0 < x 3 < , you have
4
1
1
x 3 < 4 < x
xx 3 < x 3x 0 < f x L < 2
42. f x x3
x2 9
lim f x x→3
1
6
44. (a) No. The fact that f 2 4 has no bearing on the existence of the limit of f x as x approaches 2.
3
−9
(b) No. The fact that lim f x 4 has no bearing on the
x→2
value of f at 2.
3
−3
The domain is all x ± 3. The graphing utility does not
show the hole at 3, 16 .
46. Let px be the atmospheric pressure in a plane at
altitude x (in feet).
48.
0.002
(1.999, 0.001)
(2.001, 0.001)
lim px 14.7 lbin2
x→0
1.998
0
2.002
Using the zoom and trace feature, 0.001. That is, for
0 < x 2 < 0.001,
50. True
x2 4
4 < 0.001.
x2
52. False; let
f x x10, 4x,
2
x4
x4
.
lim f x lim x2 4x 0 and f 4 10 0
x→4
x2 x 12
7
x→4
x4
54. lim
n
4 0.1
n
f 4 0.1
n
n
4 0.1
n
x→4
f 4 0.1
n
1
4.1
7.1
1
3.9
6.9
2
4.01
7.01
2
3.99
6.99
3
4.001
7.001
3
3.999
6.999
4
4.0001
7.0001
4
3.9999
6.9999
Section 1.3
56. f x mx b, m 0. Let > 0 be given. Take If 0 < x c < .
m
xc
, then
m
That is,
12L < gx L < 12L
1
2L
gx
<
3
< 2L
Hence for x in the interval c , c , x c,
1
gx > 2L > 0.
x→c
Evaluating Limits Analytically
(a) lim gx 2.4
10
x→4
(b) lim gx 4
x→0
0
4.
(a) lim f t 0
10
t→4
(b) lim f t 5
−5
t→1
10
10
−5
− 10
gx 12 x 3
x9
8. lim 3x 2 33 2 7
x→3
x→2
10. lim x2 1 12 1 0
x→1
14. lim
x→3
18. lim
x→3
2
2
2
x 2 3 2
x 1
x4
3 1
34
f t t t 4
6. lim x3 23 8
2
22. lim 2x 13 20 13 1
x→0
12. lim 3x3 2x2 4 313 212 4 5
x→1
16. lim
x→3
2x 3 23 3 3
x5
35
8
3
3
x 4 442
20. lim x→4
24. (a) lim f x 3 7 4
x→3
(b) lim gx 42 16
x→4
(c) lim g f x g4 16
x→3
26. (a) lim f x 242 34 1 21
x→4
3 21 6 3
(b) lim gx 28. lim tan x tan 0
x→ x→21
(c) lim g f x g21 3
x→4
30. lim sin
x→1
34.
1
such that 0 < x 0 < implies gx L < 2L.
which shows that lim mx b mc b.
2.
x
sin 1
2
2
lim cos x cos
x→53
5 1
3
2
309
1
58. lim gx L, L > 0. Let 2L. There exists > 0
mx c < mx mc < mx b mc b < Section 1.3
Evaluating Limits Analytically
32. lim cos 3x cos 3 1
x→ 36. lim sec
x→7
6x
sec 76 23
3
310
Chapter 1
Limits and Their Properties
38. (a) lim 4f x 4 lim f x 4
x→c
x→c
32
6
3
3
3 lim f x 27 3
f x 40. (a) lim x→c
(b) lim f x gx lim f x lim gx x→c
x→c
x→c
(c) lim f xgx lim f x lim gx x→c
x→c
x→c
x→c
lim
f x x→c f x 27 3
(b) lim
x→c 18
lim 18
18 2
3 1
2
2 2
x→c
(c) lim f x lim f x2 272 729
32
12
43
2
x→c
lim f x 32
f x
x→c
3
x→c gx
lim gx 12
x→c
(d) lim f x23 lim f x23 2723 9
(d) lim
x→c
x→c
x→c
42. f x x 3 and hx x2 3x
agree except at x 0.
x
44. gx 1
x
and f x 2
agree except at x 0.
x1
x x
(a) lim hx lim f x 5
(a) lim f x does not exist.
(b) lim hx lim f x 3
(b) lim f x 1
x→2
x→1
x→2
x→0
x→0
x→0
2x2 x 3
and gx 2x 3 agree except at
x1
x 1.
x3 1
and gx x2 x 1 agree except at
x1
x 1.
46. f x 48. f x lim f x lim gx 5
x→1
lim f x lim gx 3
x→1
x→1
x→1
7
4
−8
4
−4
50. lim
x→2
2x
x 2
lim
x2 4 x→2 x 2x 2
lim
x→2
54. lim
x
lim
x→0
x 1 2
x→3
x3
x→4
x2 5x 4
x 4x 1
lim
x2 2x 8 x→4 x 4x 2
lim
x→3
lim
x→4
2 x 2
x
x→0
lim
56. lim
52. lim
1
1
x2
4
2 x 2
x→0
4
−1
−8
2x2
2 x 2
2 x 2
2 x 2 x
x 1 2
x3
x 1 3 1
x 2 6 2
lim
x→0
1
2 x 2
x 1 2
x 1 2
lim
x→3
2
1
4
22
x3
1
1
lim
x 3x 1 2 x→3 x 1 2 4
1
1
4 x 4
x4 4
4x 4
1
1
lim
lim
58. lim
x→0
x→0
x→0 4x 4
x
x
16
60. lim
x→0
x x2 x2
x2 2x
x x2 x2
x2x x
lim
lim
lim 2x x 2x
x→0
x→0
x→0
x
x
x
Section 1.3
62. lim
x→0
Evaluating Limits Analytically
311
x x3 x3
x3 3x2
x 3x
x2 x3 x3
lim
x→0
x
x
lim
x→0
64. f x x3x2 3x
x x2
lim 3x2 3x
x x2 3x2
x→0
x
4 x
x 16
1
0
x
15.9
15.99
15.999
16
16.001
16.01
16.1
f x
.1252
.125
.125
?
.125
.125
.1248
20
−1
It appears that the limit is 0.125.
4 x
4 x
lim
x→16 x 16
x→16 x 4 x 4
Analytically, lim
lim
x→16
1
x 4
1
.
8
x5 32
80
x→2 x 2
100
66. lim
x
f x
1.9
1.99
72.39
79.20
1.999
79.92
1.9999
2.0
79.99
?
2.0001
80.01
2.001
80.08
2.01
2.1
80.80
88.41
x5 32
x 2x4 2x3 4x2 8x 16
lim
x→2 x 2
x→2
x2
Analytically, lim
lim x4 2x3 4x2 8x 16 80.
x→2
(Hint: Use long division to factor x5 32.)
68. lim
x→0
31 cos x
1 cos x
lim 3
x→0
x
x
30 0
sin x
tan2 x
sin2 x
lim
lim
2
x→0
x→0 x cos x
x→0
x
x
72. lim
cos2 x
sin x
10 0
76. lim
x→ 4
1 tan x
cos x sin x
lim
sin x cos x x→4 sin x cos x cos2 x
sin x cos x
lim
x→ 4 cos xsin x cos x
1
lim
x→ 4 cos x
lim sec x
x→ 4
2
78. lim
x→0
sin 2x
sin 2x
lim 2
sin 3x x→0
2x
13
sin3x3x
2113
1 32
70. lim
→0
cos tan sin lim
1
→0 74. lim sec 1 →
−4
3
−25
312
Chapter 1
Limits and Their Properties
80. f h 1 cos 2h
0.1
h
f h
4
0.01
1.98
0.001
1.9998
2
0
0.001
0.01
0.1
?
2
1.9998
1.98
−5
5
−4
Analytically, lim 1 cos 2h 1 cos0 1 1 2.
The limit appear to equal 2.
h→0
82. f x sin x
3 x
2
−3
0.1
x
f x
0.01
0.215
0.0464
0.001
0
0.001
0.01
0.1
0.01
?
0.01
0.0464
0.215
−2
The limit appear to equal 0.
sin x
3 2 sin x
lim x
01 0.
3
x→0 x→0
x
x
Analytically, lim
84. lim
h→0
x h x
x h x
f x h f x
lim
lim
h→0
h→0
h
h
h
lim
h→0
3
x h x
x h x
xhx
1
1
lim
h→0
h x h x
2x
x h x
f x h f x
x2 2xh h2 4x 4h x2 4x
x h2 4x h x2 4x
lim
lim
h→0
h→0
h→0
h
h
h
86. lim
lim
h→0
h2x h 4
lim 2x h 4 2x 4
h→0
h
88. lim b x a ≤ lim f x ≤ lim b x a x→a
x→a
x→a
90. f x x sin x
6
b ≤ lim f x ≤ b
x→a
Therefore, lim f x b.
x→a
− 2
2
−2
lim x sin x 0
x→0
92. f x x cos x
94. hx x cos
6
− 2
0.5
2
− 0.5
0.5
−6
− 0.5
lim x cos x 0
x→0
1
x
lim x cos
x→0
1
0
x
Section 1.3
x2 1
and gx x 1 agree at all points
x1
except x 1.
96. f x Evaluating Limits Analytically
98. If a function f is squeezed between two functions h and g,
hx ≤ f x ≤ gx, and h and g have the same limit L as
x → c, then lim f x exists and equals L.
x→c
100. f x x, gx sin2 x, hx sin2 x
x
When you are “close to” 0 the magnitude of g is “smaller”
than the magnitude of f and the magnitude of g is
approaching zero “faster” than the magnitude of f.
Thus, g f 0 when x is “close to” 0
2
g
−3
3
h
f
−2
102. st 16t2 1000 0 when t s
lim
t→5102
5 210
st
5 10
seconds
1000
16
2
510
2
t
104. 4.9t2 150 0 when t lim
t→5102
0 16t2 1000
510
t
2
16 t2 lim
t→5102
lim
t→5102
125
2
510 t
2
16 t t 5 210
t 5 10 2
16 t lim
t→5102
510
2
510
8010 ftsec 253 ftsec
2
1500
5.53 seconds.
150
4.9
49
The velocity at time t a is
sa st
4.9a2 150 4.9t2 150
4.9a ta t
lim
lim
t→a
t→a
t→a
at
at
at
lim
lim 4.9a t 2a4.9 9.8a msec.
t→a
Hence, if a 150049, the velocity is 9.8150049 54.2 msec.
106. Suppose, on the contrary, that lim gx exists. Then, since lim f x exists, so would lim f x gx, which is a
x→c
x→c
contradiction. Hence, lim gx does not exist.
x→c
x→c
108. Given f x x n, n is a positive integer, then
lim x n lim x x n1 lim xlim x n1
x→c
x→c
x→c
x→c
c lim x x n2 clim xlim x n2
x→c
x→c
cclim x→c
x→c
. . . c n.
x x n3
110. Given lim f x 0:
x→c
For every > 0, there exists > 0 such that f x 0 < whenever 0 < x c < .
Now f x 0 f x 313
f x 0 < for x c < . Therefore, lim f x 0.
x→c
314
Chapter 1
Limits and Their Properties
f x ≤ f x ≤ f x
lim f x ≤ lim f x ≤ lim f x
x→c
x→c
x→c
112. (a) If lim f x 0, then lim f x 0.
x→c
x→c
0 ≤ lim f x ≤ 0
x→c
Therefore, lim f x 0.
x→c
(b) Given lim f x L:
x→c
For every > 0, there exists > 0 such that f x L < whenever 0 < x c < .
Since f x L ≤ f x L < for x c < , then lim f x L .
x→c
116. False. Let f x 114. True. lim x3 03 0
x→0
3x xx 11
,c1
Then lim f x 1 but f 1 1.
x→1
1
118. False. Let f x 2 x2 and gx x2. Then f x < gx
for all x 0. But lim f x lim gx 0.
x→0
120. lim
x→0
1 cos x
1 cos x
lim
x→0
x
x
x→0
1 cos x
1 cos x
1 cos2 x
sin2 x
lim
x→0 x1 cos x
x →0 x1 cos x
lim
lim
x→0
sin x
x
lim
x→0
sin x
1 cos x
sin x
x
lim 1 sincosx x
x→0
10 0
122. f x sec x 1
x2
(a) The domain of f is all x 0, 2 n.
(b)
x→0
2
(d)
− 3
2
3
2
1
2
(c) lim f x sec x 1 sec x 1
x2
x2
−2
The domain is not obvious. The hole at x 0 is not
apparent.
Hence, lim
x→0
sec x 1
1 sin2 x
1
tan2 x
x2sec x 1 cos2 x x2 sec x 1
sec x 1
1 sin2 x
1
lim
2
x→0
x
cos2 x x2 sec x 1
11
124. The calculator was set in degree mode, instead of radian mode.
sec2 x 1
sec x 1 x2sec x 1
12
21.
Section 1.4
Section 1.4
2. (a)
(b)
Continuity and One-Sided Limits
315
Continuity and One-Sided Limits
lim f x 2
4. (a)
lim f x 2
(b)
x→2
x→2
lim f x 2
6. (a)
lim f x 2
(b)
x→2
x→2
lim f x 0
x→1
lim f x 2
x→1
(c) lim f x 2
(c) lim f x 2
(c) lim f x does not exist.
The function is continuous at
x 2.
The function is NOT continuous at
x 2.
The function is NOT continuous at
x 1.
x→2
8. lim
x→2
x→2
2x
1
1
lim x2 4 x→2 x 2
4
x→1
10. lim
x→4
x 2
x4
lim
x→4
lim
x→4
lim
x→4
12. lim
x→2
x 2 14. lim x→0
x2
lim
x→2
x 2
x4
x 2
x 2
x4
x 4 x 2
1
x 2
1
4
x2
1
x2
x x2 x x x2 x
x2 2xx x2 x x x2 x
lim x→0
x
x
lim x→0
2xx x2 x
x
lim 2x x 1
x→0
2x 0 1 2x 1
16. lim f x lim x2 4x 2 2
x→2
x→2
lim f x lim x→2
x→2
x2
18. lim f x lim 1 x 0
x→1
x→1
4x 6 2
lim f x 2
x→2
20. lim sec x does not exist since
x→ 2
lim
x→ 2
sec x and
lim
x→ 2
24. lim 1 x→1
x→2
sec x do not exist.
2x 1 1 2
22. lim 2x x 22 2 2
26. f x x2 1
x1
has a discontinuity at x 1 since f 1 is not defined.
x,
28. f x 2,
2x 1,
x < 1
x 1 has discontinuity at x 1 since f 1 2 lim f x 1.
x→1
x > 1
30. f t 3 9 t2 is continuous on 3, 3.
32. g2 is not defined. g is continuous on 1, 2.
316
Chapter 1
Limits and Their Properties
x
is continuous for all real x.
2
34. f x 1
is continuous for all real x.
x2 1
38. f x x
has nonremovable discontinuities at x 1 and x 1 since lim f x and lim f x do not exist.
x→1
x→1
x2 1
36. f x cos
x3
has a nonremovable discontinuity at x 3 since lim f x does not exist, and has a removable discontinuity
x→3
x2 9
at x 3 since
40. f x lim f x lim
x→3
42. f x x→3
1
1
.
x3 6
x1
x 2x 1
44. f x has a nonremovable discontinuity at x 2 since
lim f x does not exist, and has a removable discontinux→2
ity at x 1 since
lim f x lim
x→1
46. f x x→1
x 3
x3
has a nonremovable discontinuity at x 3 since lim f x
x→3
does not exist.
1
1
.
x2 3
3,
2x
x,
x < 1
x ≥ 1
2
has a possible discontinuity at x 1.
1. f 1 12 1
2.
lim f x lim 2x 3 1
x→1
x→1
x→1
lim f x 1
x→1
lim f x lim x2 1
x→1
3. f 1 lim f x
x→1
f is continuous at x 1, therefore, f is continuous for all real x.
48. f x 2x,
x 4x 1,
2
x ≤ 2
has a possible discontinuity at x 2.
x > 2
1. f 2 22 4
2.
lim f x lim 2x 4
x→2
x→2
x→2
x→2
lim f x does not exist.
x→2
lim f x lim x2 4x 1 3
Therefore, f has a nonremovable discontinuity at x 2.
50. f x csc x ,
6
2,
1. f 1 csc
2
6
2. lim f x 2
x→1
3. f 1 lim f x
x→1
x 3
x 3
≤ 2
> 2
csc x ,
6
2,
f 5 csc
1 ≤ x ≤ 5
x < 1 or x > 5
has possible discontinuities at x 1, x 5.
5
2
6
lim f x 2
x→5
f 5 lim f x
x→5
f is continuous at x 1 and x 5, therefore, f is continuous for all real x.
Section 1.4
x
has nonremovable discontinuities at each
2
2k 1, k is an integer.
58. lim g(x lim
20
x→0
x→0
lim fx 0
x→0
4 sin x
4
x
lim gx lim a 2x a
x→0
f is not continuous at x 4
x→0
−8
x→0
8
Let a 4.
−10
x2 a2
x→a x a
60. lim gx lim
x→a
lim x a 2a
x→a
Find a such that 2a 8 ⇒ a 4.
62. f gx 1
x 1
Nonremovable discontinuity at x 1. Continuous for all x > 1.
Because f g is not defined for x < 1, it is better to say that f g is discontinuous from the right at x 1.
66. hx 64. f gx sin x2
Continuous for all real x
1
x 1x 2
Nonremovable discontinuity at x 1 and x 2.
2
−3
4
−2
68. f x cos x 1 , x < 0
5x, x
lim f x lim
x→0
x→0
3
x ≥ 0
−7
f 0 50 0
cos x 1
0
x
2
−3
lim f x lim 5x 0
x→0
x→0
Therefore, lim f x 0 f 0 and f is continuous on the entire real line. (x 0 was the only possible discontinuity.)
x→0
70. f x xx 3
Continuous on 3, 317
54. f x 3 x has nonremovable discontinuities at each
integer k.
52. f x tan
56. lim f x 0
Continuity and One-Sided Limits
72. f x x1
x
Continuous on 0, 318
Chapter 1
74. f x Limits and Their Properties
x3 8
x2
76. f x x3 3x 2 is continuous on 0, 1.
f 0 2 and f 1 2
14
By the Intermediate Value Theorem, f x 0 for at least
one value of c between 0 and 1.
−4
4
0
The graph appears to be continuous on the interval
4, 4. Since f 2 is not defined, we know that f has
a discontinuity at x 2. This discontinuity is removable
so it does not show up on the graph.
78. f x 4
x
tan
is continuous on 1, 3.
x
8
f 1 4 tan
3
4
< 0 and f 3 tan
> 0.
8
3
8
By the Intermediate Value Theorem, f 1 0 for at least
one value of c between 1 and 3.
82. h 1 3 tan 80. f x x3 3x 2
f x is continuous on 0, 1.
f 0 2 and f 1 2
By the Intermediate Value Theorem, f x 0 for at least
one value of c between 0 and 1. Using a graphing utility,
we find that x 0.5961.
84. f x x2 6x 8
h is continuous on 0, 1.
f is continuous on 0, 3.
h0 1 > 0 and h1 2.67 < 0.
f 0 8 and f 3 1
By the Intermediate Value Theorem, h 0 for at least
one value between 0 and 1. Using a graphing utility, we
find that 0.4503.
1 < 0 < 8
The Intermediate Value Theorem applies.
x2 6x 8 0
x 2x 4 0
x 2 or x 4
c 2 (x 4 is not in the interval.)
Thus, f 2 0.
86. f x x2 x
x1
The Intermediate Value Theorem applies.
f is continuous on 2 , 4. The nonremovable discontinuity,
x 1, lies outside the interval.
5
f
5
35
20
and f 4 2
6
3
20
35
< 6 <
6
3
x2 x
6
x1
x2 x 6x 6
x2 5x 6 0
x 2x 3 0
x 2 or x 3
c 3 (x 2 is not in the interval.)
Thus, f 3 6.
Section 1.4
88. A discontinuity at x c is removable if you can define
(or redefine) the function at x c in such a way that the
new function is continuous at x c. Answers will vary.
(a) f x x 2
Continuity and One-Sided Limits
1,
0,
(c) f x 1,
0,
x2
sinx 2
(b) f x x2
319
if x ≥ 2
if 2 < x < 2
if x 2
if x < 2
y
3
2
1
−3
−2
−1
x
−1
1
2
3
−2
−3
90. If f and g are continuous for all real x, then so is f g (Theorem 1.11, part 2). However, fg might not be continuous if gx 0.
For example, let f x x and gx x2 1. Then f and g are continuous for all real x, but fg is not continuous at x ± 1.
1.04,
92. C 1.04 0.36t 1,
1.04 0.36t 2,
0 < t ≤ 2
t > 2, t is not an integer
t > 2, t is an integer
You can also write C as
C
Nonremovable discontinuity at each integer greater than 2.
1.04,
1.04 0.362 t,
0 < t ≤ 2
.
t > 2
C
4
3
2
1
t
1
2
3
4
94. Let st be the position function for the run up to the campsite. s0 0 (t 0 corresponds to 8:00 A.M., s20 k (distance
to campsite)). Let rt be the position function for the run back down the mountain: r0 k, r10 0. Let f t st rt.
When t 0 (8:00 A.M.),
f 0 s0 r0 0 k < 0.
When t 10 (8:10 A.M.), f 10 s10 r10 > 0.
Since f 0 < 0 and f 10 > 0, then there must be a value t in the interval 0, 10 such that f t 0. If f t 0, then
st rt 0, which gives us st rt. Therefore, at some time t, where 0 ≤ t ≤ 10, the position functions for the
run up and the run down are equal.
96. Suppose there exists x1 in a, b such that f x1 > 0 and there exists x2 in a, b such that f x2 < 0. Then by the Intermediate
Value Theorem, f x must equal zero for some value of x in x1, x2 (or x2, x1 if x2 < x1). Thus, f would have a zero in a, b,
which is a contradiction. Therefore, f x > 0 for all x in a, b or f x < 0 for all x in a, b.
98. If x 0, then f 0 0 and lim f x 0. Hence, f is
x→0
continuous at x 0.
If x 0, then lim f t 0 for x rational, whereas
t→x
lim f t lim kt kx 0 for x irrational. Hence, f is not
t →x
t →x
continuous for all x 0.
100. True
1. f c L is defined.
2. lim f x L exists.
x→c
3. f c lim f x
x→c
All of the conditions for continuity are met.
320
Chapter 1
Limits and Their Properties
102. False; a rational function can be written as PxQx
where P and Q are polynomials of degree m and n,
respectively. It can have, at most, n discontinuities.
104. (a)
S
60
50
40
30
20
10
t
5
10
15 20 25 30
(b) There appears to be a limiting speed and a possible
cause is air resistance.
106. Let y be a real number. If y 0, then x 0. If y > 0, then let 0 < x0 < 2 such that M tan x0 > y (this is possible
since the tangent function increases without bound on 0, 2). By the Intermediate Value Theorem, f x tan x is
continuous on 0, x0 and 0 < y < M, which implies that there exists x between 0 and x0 such that tan x y. The argument
is similar if y < 0.
108. 1. f c is defined.
2. lim f x lim f c x f c exists.
x→0
x→c
Let x c x. As x → c, x → 0
3. lim f x f c.
x→c
Therefore, f is continuous at x c.
110. Define f x f2x f1x. Since f1 and f2 are continuous on a, b, so is f.
f a f2a f1a > 0 and
f b f2b f1b < 0.
By the Intermediate Value Theorem, there exists c in a, b such that f c 0.
f c f2c f1c 0 ⇒ f1c f2c
Section 1.5
2.
Infinite Limits
1
x2 1
lim
x→2 x 2
lim
4.
x→2
6. f x x
x
4
lim sec
x
4
x→2
x
x2 9
3.5
f x 1.077
3.1
3.01
3.001
2.999
2.99
2.9
2.5
5.082 50.08
500.1
499.9
49.92
4.915
0.9091
lim f x x→3
lim f x x→3
lim sec
x→2
Section 1.5
8. f x sec
321
x
6
3.5
x
Infinite Limits
f x 3.864
3.1
3.01
19.11
191.0
3.001 2.999 2.99 2.9
2.5
1910
3.864
1910
191.0
19.11
lim f x x→3
lim f x x→3
10. lim
x→2
lim
x→2
4
x 23 12. lim
x→0
2x
2x
lim 2
x→0
x 1 x
x 1 x
2
Therefore, x 0 is a vertical asymptote.
4
x 23
lim
2x
x21 x lim
2x
x21 x
x→1
Therefore, x 2 is a vertical asymptote.
x→1
Therefore, x 1 is a vertical asymptote.
14. No vertical asymptote since the denominator is never zero.
16.
lim hs and lim hs .
s→5
s→5
Therefore, s 5 is a vertical asymptote.
lim hs and lim hs .
s→5
s→5
Therefore, s 5 is a vertical asymptote.
18. f x sec x x
1
has vertical asymptotes at
cos x
2n 1
, n any integer.
2
20. gx 12x3 x2 4x 1 xx2 2x 8
3x2 6x 24
6 x2 2x 8
1
x,
6
x 2, 4
No vertical asymptotes. The graph has holes at x 2
and x 4.
22. f x x
x3
4x2 x 6
4x 3x 2
4
, x 3, 2
2x2 9x 18 xx 2x2 9 xx 3
Vertical asymptotes at x 0 and x 3. The graph has holes at x 3 and x 2.
24. hx x 2x 2
x2 4
x3 2x2 x 2 x 2x2 1
has no vertical asymptote since
lim hx lim
x→2
x→2
4
x2
.
x2 1
5
26. ht tt 2
t
,t2
t 2t 2t 2 4 t 2t 2 4
Vertical asymptote at t 2. The graph has a hole at
t 2.
322
Chapter 1
28. g Limits and Their Properties
tan sin has vertical asymptotes at
cos x2 6x 7
lim x 7 8
x→1
x→1
x1
30. lim
2n 1 n, n any integer.
2
2
2
−3
3
There is no vertical asymptote at 0 since
lim
→0
tan 1.
−12
Removable discontinuity at x 1
sinx 1
1
x1
32. lim
x→1
2
34. lim
x→1
Removable discontinuity at
x 1
−3
2x
1x
3
−2
36. lim
x→4
40. lim
x→3
44.
48.
x2
x2
1
16 2
38.
x2 1
x2
9
lim
x→ 2
lim
x→ 12
lim
x→12
6x2 x 1
3x 1 5
lim
4x2 4x 3 x→12 2x 3 8
42. lim x2 x→0
2
cos x 46. lim
x→0
x2 tan x and
lim
x→ 12
x2 tan x .
Therefore, lim x2 tan x does not exist.
x→ 12
1
x
x 2
lim x 2tan x 0
x→0
cot x
50. f x x2
x3 1
x1
lim f x lim x 1 0
x→1
x→1
4
−8
8
−4
52. f x sec
x
6
54. The line x c is a vertical asymptote if the graph of f
approaches ± as x approaches c.
lim f x x→3
6
−9
9
−6
56. No. For example, f x 1
has no
x2 1
58. P vertical asymptote.
lim
V→0
k
V
k
k (In this case we know that k > 0.)
V
Section 1.5
200
ftsec
6
3
(b) r 50 sec2 200 ftsec
3
60. (a) r 50 sec2
(c)
lim
→ 2
62. m Total distance
Total time
50 2d
dx dy
50 2xy
yx
m0
lim m lim
v→c
(b)
v→c
m0
1 v2c2
x
30
40
50
60
y
150
66.667
50
42.857
(c) lim
x→25
25x
x 25 As x gets close to 25 mph, y becomes larger and larger.
50y 50x 2xy
50x 2xy 50y
50x 2yx 25
25x
y
x 25
Domain: x > 25
1
1
1
1
66. (a) A bh r2 1010 tan 102
2
2
2
2
(b)
(c)
f 50 tan 50 Domain:
0.3
0.6
0.9
1.2
1.5
0.47
4.21
18.0
68.6
630.1
0, 2 (d)
100
0
lim A → 2
1.5
0
68. False; for instance, let
f x 70. True
x2 1
.
x1
The graph of f has a hole at 1, 2, not a vertical
asymptote.
72. Let f x 1
1
and gx 4, and c 0.
x2
x
1
1
2 and lim 4 , but
x→0 x
x→0 x
lim
x1 x1 lim x x 1 0.
2
lim
x→0
2
4
x→0
4
323
1 v2c2
50 sec2 64. (a) Average speed Infinite Limits
74. Given lim f x , let g x 1. then lim
x →c
by Theorem 1.15.
x →c
gx
0
fx