ODD01 - CHAPTER 1 Limits and Their Properties Section 1.1...

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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 . . . . . . . . . . . . . . . 31 Section 1.4 Continuity and One-Sided Limits . . . . . . . . . . . . . . 37 Section 1.5 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 42 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
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27 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 tangent line at is rate of change, and equals about 0.16. x 2 5. Precalculus: sq. units Area 1 2 bh 1 2 5 3 15 2 7. Precalculus: cubic units Volume 2 4 3 24 9. (a) (b) The graphs of are approximations to the tangent line to at (c) The slope is approximately 2. For a better approximation make the list numbers smaller: 0.2, 0.1, 0.01, 0.001 x 1. y 1 y 2 8 2 4 6 (1, 3) 11. (a) (b) (c) Increase the number of line segments. 2.693 1.302 1.083 1.031 6.11 D 2 1 5 2 2 1 5 2 5 3 2 1 5 3 5 4 2 1 5 4 1 2 D 1 5 1 2 1 5 2 16 16 5.66 Section 1.2 Finding Limits Graphically and Numerically 1. Actual limit is 1 3 . lim x 2 x 2 x 2 x 2 0.3333 x 1.9 1.99 1.999 2.001 2.01 2.1 0.3448 0.3344 0.3334 0.3332 0.3322 0.3226 f x 3. Actual limit is 1 2 3 . lim x 0 x 3 3 x 0.2887 x 0.1 0.01 0.001 0.001 0.01 0.1 0.2911 0.2889 0.2887 0.2887 0.2884 0.2863 f x
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5. Actual limit is 1 16 . lim x 3 1 x 1 1 4 x 3 0.0625 x 2.9 2.99 2.999 3.001 3.01 3.1 0.0610 0.0623 0.0625 0.0625 0.0627 0.0641 f x 7. (Actual limit is 1.) (Make sure you use radian mode.) lim x 0 sin x x 1.0000 x 0.1 0.01 0.001 0.001 0.01 0.1 0.9983 0.99998 1.0000 1.0000 0.99998 0.9983 f x 9. lim x 3 4 x 1 11. lim x 2 f x lim x 2 4 x 2 15. tan x does not exist since the function increases and decreases without bound as x approaches 2. lim x 2 17. does not exist since the function oscillates between and 1 as x approaches 0. 1 lim x 0 cos 1 x (b) (c) does not exist. The values of C jump from 1.75 to 2.25 at t 3. lim t 3 C t lim t 3.5 C t 2.25 t 2 2.5 2.9 3 3.1 3.5 4 C 1.25 1.75 1.75 1.75 2.25 2.25 2.25 19. (a) 0 0 5 3 C t 0.75 0.50 t 1 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 21. You need to find such that implies That is, 1 9 > x 1 > 1 11 . 10 9 1 > x 1 > 10 11 1 10 9 > x > 10 11 9 10 < 1 x < 11 10 1 0.1 < 1 x < 1 0.1 0.1 < 1 x 1 < 0.1 f x 1 1 x 1 < 0.1. 0 < x 1 < So take Then implies Using the first series of equivalent inequalities, you obtain f x 1 1 x 1 < < 0.1. 1 11 < x 1 < 1 9 . 1 11 < x 1 < 1 11 0 < x 1 < 1 11 . 13. does not exist. For values of x to the left of 5, equals whereas for values of x to the right of 5, equals 1. x 5 x 5 1, x 5 x 5 lim x 5 x 5 x 5 28 Chapter 1 Limits and Their Properties
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23. Hence, if you have f x L < 0.01 3 x 2 8 < 0.01 3 x 6 < 0.01 3 x 2 < 0.01 0 < x 2 < 0.01 3 , 0 < x 2 < 0.01 3 0.0033 3 x 2 < 0.01 3 x 6 < 0.01 3 x 2 8 < 0.01 lim x 2 3 x 2 8 L 25. If we assume 1 < x < 3, then Hence, if you have f x L < 0.01 x 2 3 1 < 0.01 x 2 4 < 0.01 x 2 x 2 < 0.01 x 2 < 0.002 1 5 0.01 < 1 x 2 0.01 0 < x 2 < 0.002, 0.01 5 0.002. x 2 < 0.01 x 2 x 2 x 2 < 0.01 x 2 x 2 < 0.01 x 2 4 < 0.01 x 2 3 1 < 0.01 lim x 2 x 2 3 1 L 27. Given Hence, let Hence, if you have f x L < x 3 5 < x 2 < 0 < x 2 < , . x 2 < x 3 5 < > 0: lim x 2 x 3 5 29. Given Hence, let Hence, if you have f x L < 1 2 x 1 3 < 1 2 x 2 < x 4 < 2 0 < x 4 < 2 , 2 . x 4 < 2 1 2 x 4 < 1 2 x 2 < 1 2 x 1 3 < > 0: lim x 4 1 2 x 1 1 2 4 1 3 31.
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