Calculo
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 >...
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