C H A P T E R 1
Limits and Their Properties
Section 1.1
A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 27
Section 1.2
Finding Limits Graphically and Numerically . . . . . . . . 27
Section 1.3
Evaluating Limits Analytically
Section 1.4
Continuity and One-Sided Limits . . . . . . . . . . . . . . 37
Section 1.5
Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 42
. . . . . . . . . . . . . . . 31
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
C H A P T E R 1
Limits and Their Properties
Section 1.1
A Preview of Calculus
Solutions to Odd-Numbered Exercises
1. Precalculus: 20 ftsec15 seconds 300 feet
3. Calculus required: slope of tangent line at x 2 is rate of
change, and equals about 0.16.
5. Precalculus: Area 12 bh 12 53 15
2 sq. units
7. Precalculus: Volume 243 24 cubic units
9. (a)
6
(1, 3)
−4
8
−2
(b) The graphs of y2 are approximations to the tangent line to y1 at x 1.
(c) The slope is approximately 2. For a better approximation make the list numbers smaller:
0.2, 0.1, 0.01, 0.001
11. (a) D1 5 12 1 52 16 16 5.66
5
5
5
5
5
5
(b) D2 1 2 1 2 3 1 3 4 1 4 1
2
2
2
2
2.693 1.302 1.083 1.031 6.11
(c) Increase the number of line segments.
Section 1.2
1.
x
1.9
1.99
1.999
2.001
2.01
2.1
f x
0.3448
0.3344
0.3334
0.3332
0.3322
0.3226
lim
x→2
3.
Finding Limits Graphically and Numerically
x2
0.3333
x2 x 2
0.1
x
f x
lim
x→0
0.01
0.2911
0.001
0.2889
x 3 3
x
Actual limit is 13 .
0.2887
0.2887
0.001
0.01
0.1
0.2887
0.2884
0.2863
Actual limit is 1 23.
27
28
Chapter 1
5.
x
2.9
f x
0.0641
lim
x→3
7.
Limits and Their Properties
2.999
3.001
3.01
3.1
0.0627
0.0625
0.0625
0.0623
0.0610
1x 1
14
0.0625
x3
0.1
x
2.99
f x
0.01
0.9983
lim
x→0
0.001
0.99998
sin x
1.0000
x
Actual limit is 161 .
1.0000
0.001
0.01
0.1
1.0000
0.99998
0.9983
(Actual limit is 1.) (Make sure you use radian mode.)
11. lim f x lim 4 x 2
9. lim 4 x 1
x→2
x→3
13. lim
x→2
x 5 does not exist. For values of x to the left of 5, x 5x 5 equals 1,
x5
whereas for values of x to the right of 5, x 5 x 5 equals 1.
x→5
15. lim tan x does not exist since the function increases and
x→ 2
17. lim cos1x does not exist since the function oscillates
x→0
between 1 and 1 as x approaches 0.
decreases without bound as x approaches 2.
19. Ct 0.75 0.50 t 1
(a)
(b)
3
t
3
3.3
3.4
3.5
3.6
3.7
4
C 1.75
2.25
2.25
2.25
2.25
2.25
2.25
2.5
2.9
3
3.1
3.5
4
1.75
1.75
1.75
2.25
2.25
2.25
lim Ct 2.25
t→3.5
(c)
5
0
0
t
2
C 1.25
lim Ct does not exist. The values of C jump from 1.75 to 2.25 at t 3.
t→3
21. You need to find such that 0 < x 1 < implies
1
f x 1 1 < 0.1. That is,
x
0.1 <
1
1 < 0.1
x
1 0.1 <
1
x
< 1 0.1
9
<
10
1
x
11
<
10
10
>
9
x
>
10
11
10
10
1 > x 1 >
1
9
11
1
1
> x 1 > .
9
11
So take 1
. Then 0 < x 1 < implies
11
1
1
< x1 <
11
11
1
1
< x1 < .
11
9
Using the first series of equivalent inequalities, you obtain
f x 1 1
1 < < 0.1.
x
Section 1.2
23. lim 3x 2 8 L
x→2
3x 2 8
3x 6
3x 2
< 0.01
< 0.01
< 0.01
0.01
0.0033 3
0.01
Hence, if 0 < x 2 < , you have
3
0 < x2 <
3x 6 < 0.01
3x 2 8 < 0.01
f x L < 0.01
3 x 2 < 0.01
27. lim x 3 5
Finding Limits Graphically and Numerically
25. lim x2 3 1 L
x→2
x2 3 1
x2 4
x 2x 2
x 2 x 2
< 0.01
< 0.01
If we assume 1 < x < 3, then 0.015 0.002.
Hence, if 0 < x 2 < 0.002, you have
1
x 2x 2 < 0.01
x2 4 < 0.01
x2 3 1 < 0.01
f x L < 0.01
12 x 1 12 4 1 3
x 1 3 < x 2 < < 1
2
< Hence, let .
1
2
1
2
Hence, if 0 < x 2 < , you have
x2 < x 3 5 < f x L < x 4
x 4
< < 2
Hence, let 2.
Hence, if 0 < x 4 < 2, you have
x 4 < 2
x 2
x 1 3
1
2
1
2
< < f x L < 31. lim 3 3
3
33. lim x0
x→6
x→0
Given > 0:
x < 3 x 0 < Given > 0: 33 < 3
x
0 < < 3 Hence, any > 0 will work.
Hence, let 3.
Hence, for any > 0, you have
Hence for 0 < x 0 < 3, you have
1
x 2 < 0.002 50.01 < x 20.01
Given > 0:
x 3 5
x 2
< 0.01
0.01
x→4
Given > 0:
< 0.01
x 2 < x 2
29. lim
x→2
f x L < 33 < 29
x <
<
3
x 0 < f x L < 3
3 x
30
Chapter 1
Limits and Their Properties
35. lim x 2 2 2 4
x→2
37. lim x2 1 2
x→1
Given > 0:
Given > 0:
x 2 4
x 2 4
x 2 x 2 x 2
x2 1 2
x2 1
x 1x 1
< x 2 < 0
< < Hence, .
x 2 < x 2 4 < x 2 4 < f x L < If we assume 0 < x < 2, then 3.
Hence for 0 < x 1 < , you have
3
1
1
x2 1 < x2 1 2 < f x 2 < x9
0.5
41. f x x4
1
6
x 1 < 3 < x 1
(because x 20)
x 5 3
lim f x < x2 < x→4
< x 1 < x 1
Hence for 0 < x 2 < , you have
39. f x < −6
10
x 3
lim f x 6
6
x→9
−0.1667
0
10
0
The domain is 5, 4 4, .
1
The graphing utility does not show the hole at 4, 6 .
The domain is all x ≥ 0 except x 9. The graphing
utility does not show the hole at 9, 6.
43. lim f x 25 means that the values of f approach 25 as x gets closer and closer to 8.
x→8
45. (i) The values of f approach different
numbers as x approaches c from
different sides of c:
(ii) The values of f increase without bound as x approaches c:
(iii) The values of f oscillate
between two fixed numbers as
x approaches c:
y
y
y
6
4
5
3
4
2
3
4
3
2
1
1
x
−4 −3 −2 −1
−1
1
2
3
4
−3 −2 −1
−1
3
4
−3
−4
−4
47. f x 1 x1x
lim 1 x1x e 2.71828
x→0
y
7
3
(0, 2.7183)
2
1
−3 −2 −1
−1
x
1
2
2
5
−2
−3
x
−4 −3 −2
x
2
3
4
5
x
f x
x
f x
0.1
2.867972
0.1
2.593742
0.01
2.731999
0.01
2.704814
0.001
2.719642
0.001
2.716942
0.0001
2.718418
0.0001
2.718146
0.00001
2.718295
0.00001
2.718268
0.000001
2.718283
0.000001
2.718280
3
4
Section 1.3
49. False; f x sin xx is
undefined when x 0.
From Exercise 7, we have
lim
x→0
51. False; let
f x sin x
1.
x
Evaluating Limits Analytically
31
53. Answers will vary.
x10, 4x,
2
x4
.
x4
f 4 10
lim f x lim x2 4x 0 10
x→4
x→4
55. If lim f x L1 and lim f x L2, then for every > 0, there exists 1 > 0 and 2 > 0 such that x c < 1 ⇒ f x L1 < and
x→c
x→c
x c < 2 ⇒ f x L2 < . Let equal the smaller of 1 and 2. Then for x c
L1 L2 L1 f x f x L2 ≤ L1 f x f x L2 < .
Therefore, L1 L2 < 2. Since > 0 is arbitrary, it follows that L1 L2.
< , we have
57. lim f x L 0 means that for every > 0 there exists > 0 such that if
x→c
0 < x c < ,
then
f x L 0
< .
This means the same as f x L < when
0 < x c < .
Thus, lim f x L.
x→c
Section 1.3
1.
Evaluating Limits Analytically
(a) lim hx 0
7
x→5
−8
3.
(b) lim hx 6
x→0
π
−π
x→1
13
(a) lim f x 0
4
−7
(b) lim f x 0.524
−4
hx x2 5x
f x x cos x
7. lim 2x 1 20 1 1
5. lim x4 24 16
x→2
x→0
9. lim x2 3x 32 33 9 9 0
x→3
11. lim 2x2 4x 1 232 43 1 18 12 1 7
x→3
13. lim
x→2
17. lim
x→7
1 1
x
2
5x
x 2
15. lim
x→1
57
7 2
35
9
35
3
x3
13
2
2
x2 4 12 4
5
5
19. lim x 1 3 1 2
x→3
x→ 3
6 32
Chapter 1
Limits and Their Properties
21. lim x 32 4 32 1
23. (a) lim f x 5 1 4
x→4
x→1
(b) lim gx 43 64
x→4
(c) lim g f x g f 1 g4 64
x→1
25. (a) lim f x 4 1 3
27. lim sin x sin
x→ 2
x→1
(b) lim gx 3 1 2
1
2
x→3
(c) lim g f x g3 2
x→1
29. lim cos
x→2
33.
x
2
1
cos
3
3
2
lim sin x sin
x→56
31. lim sec 2x sec 0 1
x→0
5 1
6
2
35. lim tan
x→3
37. (a) lim 5gx 5 lim gx 53 15
x→c
x→c
(b) lim f x gx lim f x lim gx 2 3 5
x→c
x→c
x→c
(c) lim f xgx lim f x lim gx 23 6
x→c
x→c
x→c
lim f x
f x
2
x→c
(d) lim
x→c gx
lim gx 3
39. (a) lim f x3 lim f x3 43 64
x→c
x→c
(b) lim f x lim f x 4 2
x→c
x→c
(c) lim 3 f x 3 lim f x 34 12
x→c
x→c
(d) lim f x32 lim f x32 432 8
x→c
x→c
41. f x 2x 1 and gx x 0.
4x
tan 34 1
2x2 x
agree except at
x
x→c
43. f x xx 1 and gx x3 x
agree except at x 1.
x1
(a) lim gx lim f x 1
(a) lim gx lim f x 2
(b) lim gx lim f x 3
(b) lim gx lim f x 0
x→0
x→1
45. f x x→1
x→0
x→1
x→1
x2 1
and gx x 1 agree except at x 1.
x1
lim f x lim gx 2
x→1
x→1
47. f x x 2.
x→1
x→1
x3 8
and gx x2 2x 4 agree except at
x2
lim f x lim gx 12
x→2
3
x→2
12
−3
4
−4
−9
9
0
49. lim
x→5
x5
x5
lim
x2 25 x→5 x 5x 5
lim
x→5
1
1
x 5 10
51. lim
x→3
x2 x 6
x 3x 2
lim
x→3 x 3x 3
x2 9
lim
x→3
x 2 5 5
x 3 6 6
Section 1.3
53. lim
x 5 5
lim
x
x→0
lim
x→0
55. lim
x 5 3
x4
x→4
x 5 5
lim
x→4
x 5 5
x 5 5
5
x 5 5
1
1
lim
xx 5 5 x→0 x 5 5 25
10
x 5 3
x4
x→4
lim
x
x→0
Evaluating Limits Analytically
x 5 3
x 5 3
1
x 5 9
1
1
lim
x 4x 5 3 x→4 x 5 3 9 3 6
1
1
2 2 x
1
1
2x 2
22 x
57. lim
lim
lim
x→0
x→0
x→0 22 x
x
x
4
59. lim
x→0
2x x 2x
2x 2
x 2x
lim
lim 2 2
x→0
x→0
x
x
x x2 2x x 1 x2 2x 1
x2 2x
x x2 2x 2
x 1 x2 2x 1
lim
x→0
x→0
x
x
61. lim
lim 2x x 2 2x 2
x→0
63. lim
x 2 2
x→0
x
x
0.1
f x
0.354
0.01
0.358
2
0.001
0
0.001
0.01
0.1
0.345
?
0.354
0.353
0.349
0.354
−3
3
−2
Analytically, lim
x 2 2
x
x→0
lim
x 2 2
x
x→0
lim
x→0
x 2 2
x 2 2
x22
x x 2 2
lim
x→0
1
x 2 2
1
22
2
4
1
1
2x 2
1
65. lim
x→0
x
4
0.354
3
−5
x
0.1
0.01
0.001
0
0.001
f x
0.263
0.251
0.250
?
0.250
1
1
2x 2
2 2 x
Analytically, lim
lim
x→0
x→0
x
22 x
1
0.01
0.1
0.249
0.238
x
1
1
−2
1
1
lim
.
x x→0
lim
22 x x x→0 22 x
4
33
34
Chapter 1
67. lim
x→0
Limits and Their Properties
sin x
lim
x→0
5x
sin x
x
15
115
51
69. lim
x→0
1
sin x1 cos x
lim
x→0 2
2x2
sin x
x
1 cos x
x
1
10 0
2
1 cos h2
1 cos h
1 cos h
lim
h→0
h→0
h
h
sin2 x
sin x
lim
sin x 1 sin 0 0
x→0
x→0
x
x
71. lim
73. lim
00 0
75. lim
x→ 2
cos x
lim sin x 1
x→ 2
cot x
79. f t f t
t→0
sin 3t
sin 3t
lim
t→0
2t
3t
4
0.01
2.96
0.001
2.9996
3
0
0.001
0.01
0.1
?
3
2.9996
2.96
− 2
2
−1
The limit appear to equal 3.
sin 3t
sin 3t
lim 3
31 3.
t→0
t→0
t
3t
Analytically, lim
81. f x 32
132
23
sin 3t
t
0.1
t
77. lim
sin x2
x
1
− 2
x
0.1
0.01 0.001 0 0.001 0.01 0.1
f x
0.099998 0.01 0.001 ? 0.001 0.01 0.099998
2
−1
sin x2
sin x2
lim x
01 0.
x→0
x→0
x
x2
Analytically, lim
83. lim
h→0
f x h f x
2x h 3 2x 3
2x 2h 3 2x 3
2h
lim
lim
lim
2
h→0
h→0
h→0 h
h
h
h
4
4
4
4
f x h f x
4x 4x h
xh
x
85. lim
lim
lim
lim
2
h→0 x hx
h→0
h→0
h→0
h
h
x hxh
x
87. lim 4 x2 ≤ lim f x ≤ lim 4 x2
x→0
x→0
x→0
89. f x x cos x
4
4 ≤ lim f x ≤ 4
x→0
Therefore, lim f x 4.
x→0
− 3
2
3
2
−4
lim x cos x 0
x→0
Section 1.3
91. f x x sin x
93. f x x sin
6
2
1
x
−0.5
0.5
−6
−0.5
lim x sin x 0
lim x sin
x→0
95. We say that two functions f and g agree at all but one
point (on an open interval) if f x gx for all x in the
interval except for x c, where c is in the interval.
1
0
x
97. An indeterminant form is obtained when evaluating a limit
using direct substitution produces a meaningless fractional
expression such as 00. That is,
lim
x→c
f x
gx
for which lim f x lim gx 0
x→c
99. f x x, gx sin x, hx When you are “close to” 0 the magnitude of f is
approximately equal to the magnitude of g.
Thus, g f 1 when x is “close to” 0.
f
h
−5
x→c
sin x
x
3
g
5
−3
101. st 16t2 1000
lim
t→5
s5 st
600 16t2 1000
16t 5t 5
lim
lim
lim 16t 5 160 ftsec.
t→5
t→5
t→5
5t
5t
t 5
Speed 160 ftsec
103. st 4.9t2 150
s3 st
4.932 150 4.9t2 150
4.99 t2
lim
lim
t→3
t→3
t→3
3t
3t
3t
lim
lim
x→3
4.93 t3 t
lim 4.93 t 29.4 msec
x→3
3t
105. Let f x 1x and gx 1x. lim f x and lim gx do not exist.
x→0
x→0
lim 0 0
1
1
lim f x gx lim
x→0
x→0 x
x
x→0
107. Given f x b, show that for every > 0 there exists a > 0 such that f x b < whenever x c < . Since
f x b b b 0 < for any > 0, then any value of > 0 will work.
35
0.5
− 2
x→0
Evaluating Limits Analytically
109. If b 0, then the property is true because both sides are equal to 0. If b 0, let > 0 be given. Since lim f x L,
x→c
there exists > 0 such that f x L < b whenever 0 < x c < . Hence, wherever 0 < x c < ,
we have
b f x L
< or
bf x bL
which implies that lim bf x bL.
x→c
< 36
Chapter 1
Limits and Their Properties
M f x ≤ f xgx ≤ M f x
111.
113. False. As x approaches 0 from the left,
lim M f x ≤ lim f xgx ≤ lim M f x x→c
x→c
x→c
x 1.
x
2
M0 ≤ lim f xgx ≤ M0
x→c
−3
0 ≤ lim f xgx ≤ 0
3
x→c
Therefore, lim f xgx 0.
−2
x →c
115. True.
117. False. The limit does not exist.
4
−3
6
−2
119. Let
f x 4,4,
if x ≥ 0
if x < 0
lim f x lim 4 4.
x→0
x→0
lim f x does not exist since for x < 0, f x 4 and for x ≥ 0, f x 4.
x→0
rational
0,1, ifif xx isis irrational
0, if x is rational
g x x, if x is irrational
121. f x lim f x does not exist.
x→0
No matter how “close to” 0 x is, there are still an infinite number of rational and irrational numbers so that lim f x does not
x→0
exist.
lim gx 0.
x→0
When x is “close to” 0, both parts of the function are “close to” 0.
123. (a) lim
x→0
1 cos x
1 cos x
lim
x→0
x2
x2
lim
x→0
1 cos2 x
x 1 cos x
2
sin2 x
x→0
x2
lim
1
1 cos x
1 cos x
1
1 cos x
12
21
(b) Thus,
1 cos x 1
1
⇒ 1 cos x x2
x2
2
2
1
⇒ cos x 1 x2 for x 0.
2
1
(c) cos0.1 1 0.12 0.995
2
(d) cos0.1 0.9950, which agrees with part (c).
Section 1.4
Section 1.4
3. (a) lim f x 0
(b) lim f x 1
(b) lim f x 0
x→3
x→4
x→3
The function is continuous at
x 3.
x→0
x→4
(c) lim f x does not exist
(c) lim f x 0
x→3
x
lim
x→0 9.
does not exist because
x
x2 9
grows
x
1
1
lim
x x→0
xx x x
x→0
x→3
x
x2 9
x
1.
x
lim x→3
lim
x→3 without bound as x → 3 .
1
1
x x x
x x
x
lim 13. lim x→0
x→0
x
xx x
15. lim f x lim
The function is NOT continuous
at x 4.
The function is NOT continuous at
x 3.
x5
1
1
lim
x2 25 x→5 x 5 10
x lim f x 2
x→4 (b) lim f x 2
x→3
(c) lim f x 1
11. lim
5. (a)
x→3
x→3
x→5
37
Continuity and One-Sided Limits
1. (a) lim f x 1
7. lim
Continuity and One-Sided Limits
1
xx x
1
1
2
xx 0
x
x2 5
2
2
17. lim f x lim x 1 2
x→1
x→1
lim f x lim x3 1 2
x→1
x→1
lim f x 2
x→1
21. lim 3x 5 33 5 4
19. lim cot x does not exist since
x→ x→4
x 3 for 3 < x < 4
lim cot x and lim cot x do not exist.
x→ x→ 23. lim 2 x does not exist
x→3
because
lim2 x 2 3 5
x→3
and
25. f x 1
x2 4
27. f x has discontinuities at x 2 and
x 2 since f 2 and f 2 are not
defined.
x
x
2
has discontinuities at each integer
k since lim f x lim f x.
x→k
x→k
lim 2 x 2 4 6.
x→3
29. gx 25 x2 is continuous
on 5, 5.
31. lim f x 3 lim f x.
x→0
x→0
f is continuous on 1, 4.
33. f x x2 2x 1 is continuous
for all real x.
38
Chapter 1
Limits and Their Properties
35. f x 3x cos x is continuous for all real x.
37. f x x
is not continuous at x 0, 1. Since
x2 x
x
1
for x 0, x 0 is a removable
x2 x x 1
discontinuity, whereas x 1 is a nonremovable
discontinuity.
39. f x x
is continuous for all real x.
x2 1
41. f x x2
x 2x 5
has a nonremovable discontinuity at x 5 since lim f x
x→5
does not exist, and has a removable discontinuity at
x 2 since
lim f x lim
x→2
43. f x x 2 has a nonremovable discontinuity at x 2 since
45. f x x,x ,
x2
x→2
1
1
.
x5
7
lim f x does not exist.
x→2
x ≤ 1
x > 1
2
has a possible discontinuity at x 1.
1. f 1 1
2.
lim f x lim x 1
x→1
x→1
x→1
x→1
lim f 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.
x
1,
47. f x 2
3 x,
x ≤ 2
1. f 2 x > 2
2
12
2
lim f x lim
x→2
2.
has a possible discontinuity at x 2.
x→2
2x 1 2
lim f x lim 3 x 1
x→2
x→2
lim f x does not exist.
x→2
Therefore, f has a nonremovable discontinuity at x 2.
49. f x x
tan 4 ,
x,
x
x
1. f 1 1
2. lim f x 1
x→1
3. f 1 lim f x
x→1
x
< 1
tan 4 ,
≥ 1
x,
1 < x < 1
has possible discontinuities at x 1, x 1.
x ≤ 1 or x ≥ 1
f 1 1
lim f x 1
x→1
f 1 lim f x
x→1
f is continuous at x ± 1, therefore, f is continuous for all real x.
Section 1.4
Continuity and One-Sided Limits
39
51. f x csc 2x has nonremovable discontinuities at integer
multiples of 2.
53. f x x 1 has nonremovable discontinuities at each
integer k.
55. lim f x 0
57. f 2 8
50
x→0
lim f x 0
Find a so that lim ax2 8 ⇒ a x→0
x→2
f is not continuous at x 2.
−8
8
2.
22
8
−10
59. Find a and b such that lim ax b a b 2 and lim ax b 3a b 2.
x→1
x→3
a b 2
3a b 2
4
4a
a 1
b
2,
f x x 1,
2,
x ≤ 1
1 < x < 3
x ≥ 3
2 1 1
61. f gx x 12
63. f gx Continuous for all real x.
Nonremovable discontinuities at x ± 1
67. f x 65. y x x
Nonremovable discontinuity at each integer
0.5
1
1
x2 5 6 x2 1
2xx 2x,4,
2
x ≤ 3
x > 3
Nonremovable discontinuity at x 3
5
−3
3
−5
7
−1.5
−5
69. f x x
x2 1
71. f x sec
Continuous on , 73. f x sin x
x
Continuous on:
. . . , 6, 2, 2, 2, 2, 6, 6, 10, . . .
1
75. f x 16x4 x3 3 is continuous on 1, 2.
f 1 33
16 and f 2 4. By the Intermediate Value
Theorem, f c 0 for at least one value of c between
1 and 2.
3
−4
x
4
4
−2
The graph appears to be continuous on the interval
4, 4. Since f 0 is not defined, we know that f has
a discontinuity at x 0. This discontinuity is removable
so it does not show up on the graph.
40
Chapter 1
Limits and Their Properties
77. f x x2 2 cos x is continuous on 0, .
f 0 3 and f 2 1 > 0. By the Intermediate
Value Theorem, f c 0 for the least one value of c
between 0 and .
81. gt 2 cos t 3t
79. f x x3 x 1
f x is continuous on 0, 1.
f 0 1 and f 1 1
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 0.6823.
83. f x x2 x 1
g is continuous on 0, 1.
f is continuous on 0, 5.
g0 2 > 0 and g1 1.9 < 0.
f 0 1 and f 5 29
By the Intermediate Value Theorem, gt 0 for at least
one value c between 0 and 1. Using a graphing utility, we
find that t 0.5636.
1 < 11 < 29
The Intermediate Value Theorem applies.
x2 x 1 11
x2 x 12 0
x 4x 3 0
x 4 or x 3
c 3 (x 4 is not in the interval.)
Thus, f 3 11.
85. f x x3 x2 x 2
87. (a) The limit does not exist at x c.
f is continuous on 0, 3.
(b) The function is not defined at x c.
f 0 2 and f 3 19
(c) The limit exists at x c, but it is not equal to the
value of the function at x c.
2 < 4 < 19
The Intermediate Value Theorem applies.
(d) The limit does not exist at x c.
x3 x2 x 2 4
x3 x2 x 6 0
x 2x2 x 3 0
x2
(x2
x 3 has no real solution.)
c2
Thus, f 2 4.
89.
91. The functions agree for integer values of x:
y
5
4
3
2
1
−2 −1
gx 3 x 3 x 3 x
f x 3 x 3 x
x
1
3 4 5 6 7
−2
−3
However, for non-integer values of x, the functions
differ by 1.
f x 3 x gx 1 2 x.
The function is not continuous at x 3 because
lim f x 1 0 lim f x.
x→3
for x an integer
x→3
1
1
For example, f 2 3 0 3, g2 3 1 4.
Section 1.4
t 2 2 t
Continuity and One-Sided Limits
N
93. Nt 25 2
t
0
1
1.8
2
3
3.8
Nt
50
25
5
50
25
5
Number of units
50
40
30
20
10
t
2
Discontinuous at every positive even integer.The
company replenishes its inventory every two months.
4
6
8
10 12
Time (in months)
95. Let V 43 r 3 be the volume of a sphere of radius r.
V1 43 4.19
V5 3 53 523.6
4
Since 4.19 < 275 < 523.6, the Intermediate Value Theorem implies that there is at least one value r between 1 and 5 such
that Vr 275. (In fact, r 4.0341.)
97. Let c be any real number. Then lim f x does not exist since there are both rational and
x→c
irrational numbers arbitrarily close to c. Therefore, f is not continuous at c.
y
1, if x < 0
0, if x 0
99. sgnx 1, if x > 0
4
3
2
1
(a) lim sgnx 1
−4 −3 −2 −1
x→0
x
1
2
3
4
−2
(b) lim sgnx 1
−3
x→0
−4
(c) lim sgnx does not exist.
x→0
101. True; if f x gx, x c, then lim f x lim gx and
x→c
x→c
103. False; f 1 is not defined and lim f x does not exist.
x→1
at least one of these limits (if they exist) does not equal
the corresponding function at x c.
105. (a) f x b
0 ≤ x < b
b < x ≤ 2b
0
(b) gx y
2b
x
2
0 ≤ x ≤ b
b
x
2
b < x ≤ 2b
y
2b
b
x
b
b
2b
NOT continuous at x b.
x
b
2b
Continuous on 0, 2b.
41
42
Chapter 1
107. f x Limits and Their Properties
x c2 c
x
, c > 0
Domain: x c2 ≥ 0 ⇒ x ≥ c2 and x 0, c2, 0 0, x c2 c
lim
x
x→0
lim
x c2 c
x→0
x
x c2 c
x c2 c
x c2 c2
1
1
lim
x→0 xx c2 c
x→0 x c2 c
2c
lim
Define f 0 12c to make f continuous at x 0.
109. hx xx
15
h has nonremovable discontinuities at
x ± 1, ± 2, ± 3, . . . .
−3
3
−3
Section 1.5
1.
Infinite Limits
lim 2
x
x2 4
lim 2
x
x2 4
x→2
x→2
3.
lim tan
x
4
lim tan
x
4
x→2
x→2
5. f x 1
x2 9
x
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
f x
0.308
1.639
16.64
166.6
166.7
16.69
1.695
0.364
lim f x x→3
lim f x x→3
7. f x x2
x2 9
x
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
f x
3.769
15.75
150.8
1501
1499
149.3
14.25
2.273
lim f x x→3
lim f x x→3
Section 1.5
9. lim
x→0
1
1
lim
x2 x→0 x2
Therefore, x 0 is a vertical asymptote.
Infinite Limits
43
x2 2
x 2x 1 11. lim
x→2
x2 2
x 2x 1
lim
x→2
Therefore, x 2 is a vertical asymptote.
lim x2 2
x 2x 1 lim x2 2
x 2x 1
x→1
x→1
Therefore, x 1 is a vertical asymptote.
13.
lim x→2
x2
x2
x2
and lim 2
x→2 x 4
4
15. No vertical asymptote since the denominator is never zero.
Therefore, x 2 is a vertical asymptote.
lim
x→2
x2
x2
and lim 2
x→2 x 4
x2 4
Therefore, x 2 is a vertical asymptote.
17. f x tan 2x x
21.
lim
x→2
lim
x→2
sin 2x
has vertical asymptotes at
cos 2x
2n 1 n
, n any integer.
4
4
2
x
x 2x 1
x
x 2x 1
Therefore, x 2 is a vertical asymptote.
lim
x
x 2x 1 lim
x
x 2x 1
x→1
x→1
19. lim 1 t→0
4
4
lim 1 2
t→0
t2
t
Therefore, t 0 is a vertical asymptote.
x3 1 x 1x2 x 1
x1
x1
23. f x has no vertical asymptote since
lim f x lim x2 x 1 3
x→1
x→1
Therefore, x 1 is a vertical asymptote.
25. f x x 5x 3
x3
,x5
x 5x2 1 x2 1
No vertical asymptotes. The graph has a hole at x 5.
27. st t
has vertical asymptotes at t n, n
sin t
a nonzero integer. There is no vertical asymptote at
t 0 since
lim
t→0
t
1.
sin t
44
Chapter 1
Limits and Their Properties
x2 1
lim x 1 2
x→1 x 1
x→1
29. lim
31.
lim x2 1
x1
lim
x2 1
x1
x→1
2
−3
x→1
3
8
−3
3
Vertical asymptote at
x 1
−8
−5
Removable discontinuity at x 1
33. lim
x→2
37.
x3
x2
lim
x→3
45. lim
x→ x→3
x2 2x 3
x1 4
lim x→3 x 2
x2 x 6
5
41. lim 1 x→0
35. lim
1
x
x2 x
x
1
lim 2
x→1 x 1x 1
x→1 x 1
2
39. lim
43. lim
x→0
x
lim x sin x 0
csc x x→
x2
x 3x 3 47.
2
2
sin x lim
x→ 12
x sec x and
lim
x→ 12
x sec x .
Therefore, lim x sec x does not exist.
x→ 12
49. f x x2 x 1
x3 1
lim f x lim
x→1
x→1
51. f x 1
x1 1
x2 25
lim f x x→5
0.3
3
−8
−4
8
5
−0.3
−3
53. A limit in which f x increases or decreases without
bound as x approaches c is called an infinite limit. is
not a number. Rather, the symbol
55. One answer is f x x3
x3
.
x 6x 2 x2 4x 12
lim f x x→c
says how the limit fails to exist.
57.
k
, 0 < r < 1. Assume k 0.
1r
k
lim S lim
(or if k < 0)
r→1
r→1 1 r
59. S y
3
2
1
−2
x
−1
1
−1
−2
3
Section 1.5
61. C 528x
, 0 ≤ x < 100
100 x
63. (a) r (a) C25 $176 million
(b) r (b) C50 $528 million
(c) C75 $1584 million
x
x→25
f x
1
0.5
0.2
0.1
0.01
0.001
0.0001
0.1585
0.0411
0.0067
0.0017
0
0
0
0.5
lim
x→0
− 1.5
x sin x
0
x
1.5
− 0.25
(b)
x
f x
1
0.5
0.2
0.1
0.01
0.001
0.0001
0.1585
0.0823
0.0333
0.0167
0.0017
0
0
0.001
0.0001
0.1667 0.1667
0.1667
0.25
− 1.5
lim
1.5
x→0
x sin x
0
x2
− 0.25
(c)
x
f x
1
0.5
0.2
0.1
0.1585
0.1646
0.1663
0.1666
0.01
0.25
− 1.5
lim
1.5
x→0
x sin x
0.1167 16
x3
− 0.25
(d)
x
f x
1
0.5
0.2
0.1
0.1585
0.3292
0.8317
1.6658
0.01
0.001
0.0001
16.67
166.7
1667.0
1.5
− 1.5
1.5
− 1.5
For n ≥ 3, lim
x→0
x sin x
.
xn
lim
x→0
x sin x
x4
7
ftsec
12
215
3
ftsec
2
625 225
(c) lim
528
Thus, it is not possible.
(d) lim x→100 100 x
65. (a)
27
625 49
2x
625 x2
Infinite Limits
45
46
Chapter 1
Limits and Their Properties
(b) The direction of rotation is reversed.
67. (a) Because the circumference of the motor is
half that of the saw arbor, the saw makes
17002 850 revolutions per minute.
(d)
(c) 220 cot 210 cot : straight sections.
The angle subtended in each circle is
2 2
2
(e)
2.
0.3
0.6
0.9
1.2
1.5
L
306.2
217.9
195.9
189.6
188.5
450
Thus, the length of the belt around the pulleys is
20 2 10 2 30 2.
0
Total length 60 cot 30 2
(f)
0, 2 Domain:
2
0
lim
→ 2
L 60 188.5
(All the belts are around pulleys.)
(g) lim L →0
71. False; let
69. False; for instance, let
f x 1,
f x x
3,
x2 1
or
x1
x0
x 0.
The graph of f has a vertical asymptote at x 0, but
f 0 3.
x
gx 2
.
x 1
73. Given lim f x and lim gx L:
x→c
x→c
(2) Product:
If L > 0, then for L2 > 0 there exists 1 > 0 such that gx L < L2 whenever 0 < x c < 1. Thus,
L2 < gx < 3L2. Since lim f x then for M > 0, there exists 2 > 0 such that f x > M2L whenever
x c
x→c
< 2. Let be the smaller of 1 and 2. Then for 0 < x c < , we have f xgx > M2LL2 M.
Therefore lim f xgx . The proof is similar for L < 0.
x→c
(3) Quotient: Let > 0 be given.
There exists 1 > 0 such that f x > 3L2
whenever 0 < x c < 1 and there exists 2 > 0 such that gx L <
L2 whenever 0 < x c < 2. This inequality gives us L2 < gx < 3L2. Let be the smaller of 1 and 2. Then
for 0 <
x c
gx
f x
<
3L2
.
3L2
Therefore, lim
x→c
75. Given lim
x→c
< , we have
gx
0.
f x
1
0.
f x
Suppose lim f x exists and equals L. Then,
x→c
lim 1
1
1
x→c
0.
x→c f x
lim f x L
lim
x→c
This is not possible. Thus, lim f x does not exist.
x→c