Evnrev01

324

Chapter 1

Limits and Their Properties

Review Exercises for Chapter 1 2. Precalculus. L 9 1 2 3 1 2 8.25 4.

0.1

x f x

0.01

0.358

0.5

0.001

0.001

0.01

0.1

0.354

0.354

0.353

0.349

0.354

−1

1

lim f x 0.2

x→0

6. g x

−0.5

3x x 2

(b) lim g x 0

(a) lim g x does not exist.

x→0

x→2

Assuming 4 < x < 16, you can choose 5 .

8. lim x 9 3. x→9

Let > 0 be given. We need

x→5

x 9 < x 3

x 9 < 5 < x 3

x 3 < ⇒ x 3 x 3 < x 3

10. lim 9 9. Let > 0 be given. can be any positive

Hence, for 0 < x 9 < 5 , you have

x 3 <

f x L <

12. lim 3 y 1 3 4 1 9 y→4

number. Hence, for 0 < x 5 < , you have

9 9 < f x L < 14. lim t→3

t2 9 lim t 3 6 t→3 t 3

16. lim

4 x 2

x→0

x

lim lim

x→0

18. lim

s→0

x 1 4 x 2

1 1 s 1 lim 1 1 s 1 1 1 s 1

s→0

s s 1 1 s 1 lim

s→0

20. lim

4 x 2

x→0

x→ 2

1 1 s 1 1 1 lim s 1 1 s 1 s→0 1 s 1 1 s 1 2

x2 4 x 2 x 2 lim x3 8 x→ 2 x 2 x2 2x 4 lim

x→ 2

x 2 x2 2x 4

4 1

12 3

22.

lim

x→ 4

4x 4 4

tan x 1

4 x 2 4 x 2

1 4

Review Exercises for Chapter 1

24. lim

xâ&#x2020;&#x2019;0

cos x 1 cos cos x sin sin x 1 lim xâ&#x2020;&#x2019;0 x x lim

xâ&#x2020;&#x2019;0

cos x x 1

lim sin

xâ&#x2020;&#x2019;0

sin x x

0 0 1 0 3 2 7 26. lim f x 2g x 4 2 3 12 xâ&#x2020;&#x2019;c

28. f x (a)

3 x 1 x 1

x f x

lim

xâ&#x2020;&#x2019;1

(c) lim xâ&#x2020;&#x2019;1

1.1

1.01

1.001

0.3228

0.3322

0.3332

3 x 1 0.333 x 1

0.3333

3

â&#x2C6;&#x2019;3

3 x 1 3 x 3 3 x 2 1 x

1 x 3 3 x 1 1 x x 2

lim

1 3 x 3 x 2 1

â&#x2C6;&#x2019;3

2

lim

xâ&#x2020;&#x2019;1

2

Actual limit is 13 .

3 x 3 x 1 1 lim xâ&#x2020;&#x2019;1 x 1 x 1

xâ&#x2020;&#x2019;1

(b)

1.0001

1 3

30. s t 0 â&#x2021;&#x2019; 4.9t2 200 0 â&#x2021;&#x2019; t2 40.816 â&#x2021;&#x2019; t 6.39 sec When t 6.39, the velocity is approximately lim tâ&#x2020;&#x2019;a

s a s t lim 4.9 a t tâ&#x2020;&#x2019;a a t lim 4.9 6.39 6.39 62.6 m sec. tâ&#x2020;&#x2019;6.39

32. lim x 1 does not exist. The graph jumps from 2 to 3 xâ&#x2020;&#x2019;4

34. lim g x 1 1 2. xâ&#x2020;&#x2019;1

at x 4.

36. lim f s 2 sâ&#x2020;&#x2019; 2

38. f x

3x2 x 2 , x 1 0,

lim f x lim

xâ&#x2020;&#x2019;1

xâ&#x2020;&#x2019;1

x 1 x 1

3x x 2 x 1 2

lim 3x 2 5 0 xâ&#x2020;&#x2019;1

Removable discontinuity at x 1 Continuous on , 1 1,

325

326

Chapter 1

52x x,3,

40. f x

Limits and Their Properties

x x 1 1 1x 1 lim 1

x

x â&#x2030;¤ 2 x > 2

42. f x

lim 5 x 3

xâ&#x2020;&#x2019;2

xâ&#x2020;&#x2019;0

lim 2x 3 1

Domain: , 1 , 0,

xâ&#x2020;&#x2019;2

Nonremovable discontinuity at x 2 Continuous on , 2 2,

Nonremovable discontinuity at x 0 Continuous on , 1 0,

x 1 2x 2

44. f x

46. f x tan 2x Nonremovable discontinuities when

x 1 1

xâ&#x2020;&#x2019; 1 2 x 1 2 lim

x

Removable discontinuity at x 1 Continuous on , 1 1,

2n 1 4

Continuous on

2n 4 1 , 2n 4 1 for all integers n.

48. lim x 1 2 xâ&#x2020;&#x2019;1

lim x 1 4

xâ&#x2020;&#x2019;3

Find b and c so that lim x2 bx c 2 and lim x2 bx c 4. xâ&#x2020;&#x2019;1

xâ&#x2020;&#x2019;3

1 b c 2

Consequently we get Solving simultaneously,

b

and 9 3b c 4.

3 and

50. C 9.80 2.50 x 1 , x > 0

c 4.

52. f x x 1 x (a) Domain: , 0 1,

9.80 2.50 x 1

(b) lim f x 0

C has a nonremovable discontinuity at each integer.

xâ&#x2020;&#x2019;0

30

(c) lim f x 0 xâ&#x2020;&#x2019;1

0

5 0

54. h x

4x 4 x2

56. f x csc x Vertical asymptote at every integer k

Vertical asymptotes at x 2 and x 2

58.

62.

lim

xâ&#x2020;&#x2019; 1 2

lim

xâ&#x2020;&#x2019; 1

66. lim xâ&#x2020;&#x2019;0

x

2x 1

x2 2x 1

x 1

sec x

x

60.

lim

xâ&#x2020;&#x2019; 1

64. lim xâ&#x2020;&#x2019;2

68. lim xâ&#x2020;&#x2019;0

x 1 1 1

lim x4 1 xâ&#x2020;&#x2019; 1 x2 1 x 1 4 1

3 2

x 4

cos2 x

x

Problem Solving for Chapter 1 tan 2x x

70. f x (a)

327

0.1

x f x lim

x→0

0.01

2.0271

0.001

2.0003

2.0000

0.001

0.01

0.1

2.0000

2.0003

2.0271

tan 2x 2 x

(b) Yes, define f x

2, x

tan 2x ,

x 0 x 0

.

Now f x is continuous at x 0.

Problem Solving for Chapter 1 1 1 x 2. (a) Area PAO bh 1 x 2 2 2

4. (a) Slope

1 1 y x2 Area PBO bh 1 y 2 2 2 2 (b) a x

4 0 4

3 0 3

(b) Slope

3 Tangent line: y 4 x 3 4

3 4

x2 2 Area PBO

x Area PAO x 2

3 25 y x 4 4

x

4

2

1

0.1

0.01

Area PAO

2

1

1 2

1 20

1 200

Area PBO

8

2

1 2

1 200

1 20,000

a x

4

2

1

1 10

(c) Let Q x, y x, 25 x2 mx

x 3

(d) lim mx lim x→3

1 100

x→3

25 x2 4

x 3

25 x2 4 25 x2 4

25 x2 16 x→3 x 3 25 x2 4

lim

(c) lim a x lim x 0 x→0

25 x2 4

x→0

lim

x→3

lim

x→3

3 x 3 x x 3 25 x2 4 3 x 25 x2 4

3 6

4 4 4

This is the slope of the tangent line at P.

6.

a bx 3

x

a bx 3

x

a bx 3 a bx 3

x

x→0

Setting

b 3 3

b 3 bx 3

3, you obtain b 6.

Thus, a 3 and b 6.

ax tan x a because lim 1 x→0 tan x x

a2 2 a

bx lim x→0 x 3 bx 3 lim

x→0

Thus,

Thus, lim

x→0

x→0

Letting a 3 simplifies the numerator.

x→0

x→0

lim f x lim

a bx 3 x a bx 3

3 bx 3

8. lim f x lim a2 2 a2 2

.

a2 a 2 0

a 2 a 1 0 a 1, 2

328

Chapter 1

Limits and Their Properties 1 (a) f 4 4 4

y

10.

f 3

1 3

3 2

0

f 1 1 1

1 x

−1

1

−1

(b) lim f x 1

(c) f is continuous for all real numbers except

x→1

lim f x 0

x→1

x 0, ± 1, ± 12, ± 13, . . .

lim f x

x→0

lim f x

x→0

−2

v2

12. (a)

192,000 v02 48 r

192,000 v2 v02 48 r r lim r

v→0

192,000 v v02 48 192,000 48 v02

Let v0 48 4 3 feet sec. (b)

v2

1920 v02 2.17 r

1920 v2 v02 2.17 r r lim r

v→0

1920 v2 v02 2.17 1920 2.17 v02

Let v0 2.17 mi sec 1.47 mi sec . r

(c)

lim r

v→0

10,600 v2 v02 6.99 10,600 6.99 v02

Let v0 6.99 2.64 mi sec.

Since this is smaller than the escape velocity for earth, the mass is less.

14. Let a 0 and let > 0 be given. There exists 1 > 0 such that if 0 < x 0 < , then f x L < .

Let

1 a . Then for 0 < x 0 <

1 a , you have

x < a1

ax < 1

f ax L < .

As a counterexample, let f x Then lim f x 1 L, x→0

but lim f ax lim f 0 2. x→0

x→0

12

x 0 . x 0