# Calculo

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C H A P T E R 1 Limits and Their Properties
Section 1.1 Section 1.2 Section 1.3 Section 1.4 Section 1.5 A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 27 Finding Limits Graphically and Numerically . . . . . . . . 27 Evaluating Limits Analytically . . . . . . . . . . . . . . . 31

Continuity and One-Sided Limits . . . . . . . . . . . . . . 37 Infinite Limits . . . . . . . . . . .. . . . . . . . . . . . . 42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Review Exercises Problem Solving

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 ft sec 15 seconds 300 feet 3. Calculus required: slope of tangentline at x change, and equals about 0.16. 2 is rate of

5. Precalculus: Area

1 2 bh

1 2

5 3

15 2 sq.

units

7. Precalculus: Volume

2 4 3

24 cubic units

9. (a)

6

(1, 3)
−4 −2 8

(b) The graphs of y2 are approximations to the tangent line to y1 at x 0.2, 0.1, 0.01, 0.001 11. (a) D1 (b) D2 5 1 2.693 1
2

1.

(c) The slope is approximately 2. For a betterapproximation make the list numbers smaller:

1 1

5

2

16
5 2 5 2 3

16 1 6.11

5.66
5 3 5 2 4

5 2 2

1

5 4

1

2

1.302

1.083

1.031

(c) Increase the number of line segments.

Section 1.2
1. x f x lim 1.9 0.3448 x x2 x 2

Finding Limits Graphically and Numerically
1.99 0.3344 1.999 0.3334 2.001 0.3332 2.01 0.3322 2.1 0.3226

x→2

2

0.3333

Actuallimit is 1 . 3

3.

x f x lim

0.1 0.2911 x 3 x

0.01 0.2889 3 0.2887

0.001 0.2887

0.001 0.2887

0.01 0.2884

0.1 0.2863

x→0

Actual limit is 1 2 3 .

27

28 5.

Chapter 1

Limits and Their Properties

x f x

2.9 0.0641 1 x x 1 3

2.99 0.0627 1 4

2.999 0.0625

3.001 0.0625

3.01 0.0623
1 16 .

3.1 0.0610

x→3

lim

0.0625

Actual limit is

7.x f x lim

0.1 0.9983 sin x x x 1 1.0000

0.01 0.99998

0.001 1.0000

0.001 1.0000

0.01 0.99998

0.1 0.9983

x→0

(Actual limit is 1.) (Make sure you use radian mode.)

9. lim 4
x→3

11. lim f x
x→2

x→2

lim 4

x

2

13. lim

x 5 does not exist. For values of x to the left of 5, x 5 x x 5 whereas for values of x to the right of 5, x 5 x 5 equals 1.
x→5

5equals

1,

15. lim tan x does not exist since the function increases and
x→ 2

17. lim cos 1 x does not exist since the function oscillates
x→0

decreases without bound as x approaches

2. (b)

between

1 and 1 as x approaches 0.

19. C t (a)
3

0.75

0.50

t

1

t

3

3.3 2.25 2.25 2.5 1.75

3.4 2.25

3.5 2.25

3.6 2.25

3.7 2.25

4 2.25

C 1.75 lim Ct t 2

t→3.5
0 0 5

(c)

2.9 1.75

3 1.75

3.1 2.25

3.5 2.25

4 2.25 3.

C 1.25

lim C t does not exist. The values of C jump from 1.75 to 2.25 at t
t→3

21. You need to find 1 f x 1 x 0.1 < 1 0.1 < 9 < 10 10 > 9 10 9 1 > x 1 > x 9 1 x

such that 0 < x 1 < 0.1. That is, 1 < 0.1 1 x 1 x x
< 1

1 <

implies

So take

1 . Then 0 < x 11 1 < 1 11

1 <

implies

1< x 11 1 < x 11

1 1 < . 9

0.1

Using the first series of equivalent inequalities, you obtain f x 1 1 x 1 <
< 0.1.

11 < 10
>

10 11 10 11 1 . 11 1

1 > 1 >

Section 1.2 23. lim 3x
x→2

Finding Limits Graphically and Numerically 3 3 x x
2

29

2 2

8

L

25. lim x2
x→2

1

L

3x

8 < 0.01 6 < 0.01 2 < 0.01 2 < 0.01 3 2 < 2 < 0.01 6 < 0.01 8 < 0.01 L < 0.010.0033 0.01 , you have 3

x2

1 < 0.01 4 < 0.01 2 < 0.01

3x 3x 0 < x

2 x 2 x x

x

2 < 0.01 2 < 0.01 x 2 0.01 5 0.002.

Hence, if 0 < x 3x 3x 3x 2 f x

If we assume 1 < x < 3, then Hence, if 0 < x x x 2 x x2 x2 3 f x 2 < 2 < 0.002 2 < 0.01 4 < 0.01 1 < 0.01 L < 0.01

0.002, you have 1 1 0.01 < 0.01 5 x 2

27. lim x
x→2

3 > 0: 3 x

5

29. lim

x→ 4

1 2x

1 >...