Even01

Page 1

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 ft sec 15 seconds 300 feet 6. Precalculus: Area 2

8. Precalculus: Volume 3 26 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.

x 2 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

x 3

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

x x 1 4 5 0.04 x 4 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 x 3 function increases and decreases without bound as x approaches 3.

x→1

16. lim sec x 1

14. lim

x→3

18. lim sin x 0

x→0

x→1

20. C t 0.35 0.12 t 1

(a)

1

0

5 0

(b)

t C t

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 C t 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 C t

lim C t 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 < x 2

< < < < <

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 < 2 x 3 <

x 5 x 5 < 0.01 0.01

x 5 < x 5

x 3

If we assume 4 < x < 6, then 0.01 11 0.0009.

0.01 , you have 11

x 5 <

0.01 1 < 0.01 11 x 5

Hence, if 0 < x 5 <

x 5 x 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 3 2 .

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 x x 3

x2

x 3 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

x x 3 < x 3x 0 < f x L < 2

42. f x

x 3 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 Âą 3. The graphing utility does not show the hole at 3, 16 . 46. Let p x be the atmospheric pressure in a plane at altitude x (in feet).

48.

0.002

(1.999, 0.001) (2.001, 0.001)

lim p x 14.7 lb in2

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. x 2

52. False; let f x

x10, 4x, 2

x 4 x 4

.

lim f x lim x2 4x 0 and f 4 10 0

x→4

x2 x 12 7 x→4 x 4

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 < g x L < 12L 1 2L

g x

<

3

< 2L

Hence for x in the interval c , c , x c, 1 g x > 2L > 0.

x→c

Evaluating Limits Analytically (a) lim g x 2.4

10

x→4

(b) lim g x 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

g x

12 x 3 x 9

8. lim 3x 2 3 3 2 7 x→ 3

x→ 2

10. lim x2 1 1 2 1 0 x→1

14. lim

x→ 3

18. lim

x→3

2 2 2 x 2 3 2

x 1

x 4

3 1

3 4

f t t t 4

6. lim x3 2 3 8

2

22. lim 2x 1 3 2 0 1 3 1 x→0

12. lim 3x3 2x2 4 3 1 3 2 1 2 4 5 x→1

16. lim

x→3

2x 3 2 3 3 3 x 5 3 5 8

3 3 x 4 4 4 2 20. lim x→4

24. (a) lim f x 3 7 4 x→ 3

(b) lim g x 42 16 x→4

(c) lim g f x g 4 16 x→ 3

26. (a) lim f x 2 42 3 4 1 21 x→4

3 21 6 3 (b) lim g x

28. lim tan x tan 0 x→

x→21

(c) lim g f x g 21 3 x→4

30. lim sin x→1

34.

1 such that 0 < x 0 < implies g x L < 2L.

which shows that lim mx b mc b.

2.

x sin 1 2 2

lim cos x cos

x→5 3

5 1 3 2

309

1 58. lim g x L, L > 0. Let 2L. There exists > 0

m x 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 g x lim f x lim g x x→c

x→c

x→c

(c) lim f x g x lim f x lim g x 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 x 2 27 2 729

32 12 43

2

x→c

lim f x 3 2 f x x→c 3 x→c g x lim g x 1 2

x→c

(d) lim f x 2 3 lim f x 2 3 27 2 3 9

(d) lim

x→c

x→c

x→c

42. f x x 3 and h x

x2 3x agree except at x 0. x

44. g x

1 x and f x 2 agree except at x 0. x 1 x x

(a) lim h x lim f x 5

(a) lim f x does not exist.

(b) lim h x 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 g x 2x 3 agree except at x 1 x 1.

x3 1 and g x x2 x 1 agree except at x 1 x 1.

46. f x

48. f x

lim f x lim g x 5

x→ 1

lim f x lim g x 3

x→ 1

x→ 1

x→ 1

7

4

−8

4

−4

50. lim

x→2

2 x x 2 lim x2 4 x→2 x 2 x 2 lim

x→2

54. lim

x

lim

x→0

x 1 2

x→3

x 3

x→4

x2 5x 4 x 4 x 1 lim x2 2x 8 x→4 x 4 x 2

lim

x→3

lim

x→4

2 x 2

x

x→0

lim

56. lim

52. lim

1 1 x 2 4

2 x 2

x→0

4 −1

−8

2 x 2

2 x 2 2 x 2

2 x 2 x

x 1 2

x 3

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 2 2

x 3 1 1 lim x 3 x 1 2 x→3 x 1 2 4

1 1 4 x 4 x 4 4 4 x 4 1 1 lim lim 58. lim x→0 x→0 x→0 4 x 4 x x 16

60. lim

x→0

x x 2 x2 x2 2x x x 2 x2

x 2x 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 x 3 x3 x3 3x2 x 3x x 2 x 3 x3 lim

x→0

x

x lim

x→0

64. f x

x 3x2 3x x x 2 lim 3x2 3x x x 2 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 2 x4 2x3 4x2 8x 16 lim x→2 x 2 x→2 x 2

Analytically, lim

lim x4 2x3 4x2 8x 16 80. x→2

(Hint: Use long division to factor x5 32.)

68. lim

x→0

3 1 cos x 1 cos x lim 3 x→0 x x

3 0 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

1 0 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 x sin 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 2 1 13 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 cos 0 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 0 1 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

x h x 1 1 lim h→0 h x h x 2 x x h x

f x h f x x2 2xh h2 4x 4h x2 4x x h 2 4 x h x2 4x lim lim h→0 h→0 h→0 h h h

86. lim

lim

h→0

h 2x 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. h x 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 g x x 1 agree at all points x 1 except x 1.

96. f x

Evaluating Limits Analytically

98. If a function f is squeezed between two functions h and g, h x ≤ f x ≤ g x , 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, g x sin2 x, h x

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. s t 16t2 1000 0 when t s lim

t→5 10 2

5 210 s t

5 10 seconds 1000 16 2

5 10 2

t

104. 4.9t2 150 0 when t

lim

t→5 10 2

0 16t2 1000 5 10 t 2

16 t2 lim

t→5 10 2

lim

t→5 10 2

125 2

5 10 t 2

16 t

t 5 210

t 5 10

2

16 t lim

t→5 10 2

5 10 2

5 10 80 10 ft sec 253 ft sec 2

1500 5.53 seconds. 150 4.9 49

The velocity at time t a is s a s t 4.9a2 150 4.9t2 150 4.9 a t a t lim lim t→a t→a t→a a t a t a t

lim

lim 4.9 a t 2a 4.9 9.8a m sec. t→a

Hence, if a 1500 49, the velocity is 9.8 1500 49 54.2 m sec. 106. Suppose, on the contrary, that lim g x exists. Then, since lim f x exists, so would lim f x g x , which is a x→c

x→c

contradiction. Hence, lim g x 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 n 1 lim x lim x n 1

x→c

x→c

x→c

x→c

c lim x x n 2 c lim x lim x n 2 x→c

x→c

c c lim x→c

x→c

. . . c n.

x x n 3

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

,c 1

Then lim f x 1 but f 1 1. x→1

1 118. False. Let f x 2 x2 and g x x2. Then f x < g x for all x 0. But lim f x lim g x 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 x 1 cos x x →0 x 1 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

1 0 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 x2 sec 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

1 1

124. The calculator was set in degree mode, instead of radian mode.

sec2 x 1

sec x 1 x2 sec 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

2 x 1 1 lim x2 4 x→2 x 2 4

x→ 1

10. lim x→4

x 2

x 4

lim x→4

lim x→4

lim x→4

12. lim x→2

x 2

14. lim x→0

x 2

lim

x→2

x 2

x 4

x 2 x 2

x 4 x 4 x 2 1 x 2

1 4

x 2 1 x 2

x x 2 x x x2 x x2 2x x x 2 x x x2 x lim x→0 x x lim x→0

2x x x 2 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 2 2 2 2

26. f x

x2 1 x 1

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. g 2 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

x 3 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 . x 3 6

x 1 x 2 x 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

x 3 has a nonremovable discontinuity at x 3 since lim f x x→3 does not exist.

1 1 . x 2 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 2 2 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 f x 0

x→0

4 sin x 4 x

lim g x 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 g x lim x→a

lim x a 2a x→a

Find a such that 2a 8 ⇒ a 4.

62. f g x

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. h x

64. f g x sin x2 Continuous for all real x

1 x 1 x 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 5 0 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 x x 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

x 1 x

Continuous on 0,

318

Chapter 1

74. f x

Limits and Their Properties

x3 8 x 2

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 .

h 0 1 > 0 and h 1 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 2 x 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 x 1

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 x 1 x2 x 6x 6 x2 5x 6 0

x 2 x 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,

x 2 sin x 2 (b) f x x 2

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, f g might not be continuous if g x 0. For example, let f x x and g x x2 1. Then f and g are continuous for all real x, but f g is not continuous at x Âą 1.

1.04, 92. C 1.04 0.36 t 1 , 1.04 0.36 t 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 0.36 2 t ,

0 < t ≤ 2 . t > 2

C 4 3 2 1 t 1

2

3

4

94. Let s t be the position function for the run up to the campsite. s 0 0 (t 0 corresponds to 8:00 A.M., s 20 k (distance to campsite)). Let r t be the position function for the run back down the mountain: r 0 k, r 10 0. Let f t s t r t . When t 0 (8:00 A.M.),

f 0 s 0 r 0 0 k < 0.

When t 10 (8:10 A.M.), f 10 s 10 r 10 > 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 s t r t 0, which gives us s t r t . 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 P x Q x 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 f2 x f1 x . Since f1 and f2 are continuous on a, b , so is f. f a f2 a f1 a > 0 and

f b f2 b f1 b < 0.

By the Intermediate Value Theorem, there exists c in a, b such that f c 0. f c f2 c f1 c 0 ⇒ f1 c f2 c

Section 1.5 2.

Infinite Limits

1 x 2 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 2 3

12. lim x→0

2 x 2 x lim 2 x→0 x 1 x x 1 x 2

Therefore, x 0 is a vertical asymptote.

4 x 2 3

lim

2 x x2 1 x

lim

2 x x2 1 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 h s and lim h s .

s→ 5

s→ 5

Therefore, s 5 is a vertical asymptote. lim h s and lim h s .

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. g x

1 2 x3 x2 4x 1 x x2 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

4 x2 x 6 4 x 3 x 2 4 , x 3, 2 2x2 9x 18 x x 2 x2 9 x x 3

Vertical asymptotes at x 0 and x 3. The graph has holes at x 3 and x 2.

24. h x

x 2 x 2 x2 4 x3 2x2 x 2 x 2 x2 1

has no vertical asymptote since lim h x lim

x→ 2

x→ 2

4 x 2 . x2 1 5

26. h t

t t 2 t ,t 2 t 2 t 2 t 2 4 t 2 t 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 x 1

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

sin x 1 1 x 1

32. lim

x→ 1

2

34. lim x→1

Removable discontinuity at x 1

−3

2 x 1 x

3

−2

36. lim x→4

40. lim

x→3

44.

48.

x2

x2 1 16 2

38.

x 2 1 x2 9

lim

x→ 2

lim

x→ 1 2

lim

x→ 1 2

6x2 x 1 3x 1 5 lim 4x2 4x 3 x→ 1 2 2x 3 8

42. lim x2 x→0

2 cos x

46. lim

x→0

x2 tan x and

lim

x→ 1 2

x2 tan x .

Therefore, lim x2 tan x does not exist. x→ 1 2

1 x

x 2 lim x 2 tan x 0 x→0 cot x

50. f x

x2

x3 1 x 1

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 ft sec 6 3 (b) r 50 sec2 200 ft sec 3

60. (a) r 50 sec2

(c)

lim

→ 2

62. m

Total distance Total time

50

2d d x d y

50

2xy y x

m0

lim m lim

v→c

(b)

v→c

m0 1 v2 c2

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 2y x 25 25x y x 25 Domain: x > 25 1 1 1 1 66. (a) A bh r2 10 10 tan 10 2 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 . x 1

The graph of f has a hole at 1, 2 , not a vertical asymptote.

72. Let f x

1 1 and g x 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 v2 c2

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

g x 0 f x

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