ODDREV01 - Review Exercises for Chapter 1 47 Review...

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Unformatted text preview: Review Exercises for Chapter 1 47 Review Exercises for Chapter 1 1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3. Or, the length is slightly longer than the distance between the two points, 8.25. 3. x fx lim f x 0.1 0.26 0.25 0.01 0.25 0.001 0.250 0.001 0.2499 0.01 0.249 0.1 0.24 −1 1 1 x→0 −1 5. h x x2 x 2x (a) lim h x x→0 2 3 7. lim 3 x→1 x 3 1 2 . Then for Let (b) lim h x x→ 1 > 0 be given. Choose 0< x 1< x 1 1< x< 2< L< , you have 3 x fx 9. lim x2 x→2 3 1 3 1< ⇒ x2 4 x 2x 2< 2 < Let > 0 be given. We need x2 ⇒x 2< 1 x 2 . Assuming, 1 < x < 3, you can choose x x 2x x2 x2 3 fx 11. lim t→4 5. Hence, for 0 < x 5 you have 2< 2< 4< 1< L< 4 5 < 1 x 2 t 2 2 6 2.45 13. lim t→ 2 t t2 2 4 t→ 2 lim 1 t 2 1 4 15. lim x x x→4 2 4 x→4 lim x 1 x x x 2 2 2 x 2 1 4 25 17. lim 1x x→0 1 x 1 x→4 lim 1 42 5 x2 x 5x 5x 5 25 x1 xx 1 1 1 lim x→0 x 1 x→0 lim 1 19. lim x→ x3 5x 125 5 x→ 5 lim 21. lim 1 x→0 cos x sin x x→0 lim x sin x 1 cos x x 10 0 x→ 5 lim x2 75 48 Chapter 1 sin 6 Limits and Their Properties x x 12 sin 1 2 6 cos x cos x x 3 2 1 cos x lim x→0 23. lim x→0 lim lim 0 6 sin x 3 2 sin x x 12 x→0 x→0 3 1 2 25. lim f x x→c gx 2x x 1 1 3 4 2 3 1 2 27. f x (a) x fx 3 1.1 0.5680 2x x 2x x 1 1 1 1 1.01 0.5764 3 3 1.001 0.5773 1.0001 0.5773 (b) 2 x→1 lim 0.577 lim Actual limit is 2x x 2x 1 1 1 3 2x 1 3 3 3 4.9t2 t 3 3 3. 2x 2x 3 1 1 3 3 −1 0 2 (c) lim x→1 x→1 x→1 lim lim x 2x 1 2 1 x→1 2 23 sa t→a a st t 4.9 4 4.9 t 4 4.9 t 2 1 3 200 4 4 29. lim lim t→4 200 31. lim x→3 x x 3 3 x→3 lim x x 3 3 1 lim t→4 4t t 4 lim t→4 39.2 m sec 33. lim f x x→2 0 35. lim h t does not exist because lim h t t→1 t→1 t→1 1 1 2 and lim h t 1 2 1 1 1. 37. f x x→k x→k x 3 3 3 k k 3 where k is an integer. 2 where k is an integer. 39. f x x→1 3x2 x x 1 2 2 3x x 5 lim x lim x 2x 1 1 lim f x x→1 lim 3x Nonremovable discontinuity at each integer k Continuous on k, k 1 for all integers k 1 x 1 x 2 2 Removable discontinuity at x Continuous on ,1 1, 3 x 1 1 41. f x lim 2 2 43. f x x→1 x→1 lim f x x→2 lim f x 1 Nonremovable discontinuity at x Continuous on ,2 2, 2 Nonremovable discontinuity at x Continuous on ,1 1, Problem Solving for Chapter 1 45. f x csc x 2 47. f 2 5 x→2 49 Nonremovable discontinuities at each even integer. Continuous on 2k, 2k 2 Find c so that lim cx c2 6 2c c 5 1 1 2 4 2 6 5. for all integers k. 1 < 0 and 49. f is continuous on 1, 2 . f 1 f2 13 > 0. Therefore by the Intermediate Value Theorem, there is at least one value c in 1, 2 such that 2c3 3 0. 51. f x x→2 x2 x x 4 4 2 x x 2 2 (a) lim f x (b) lim f x x→2 x→2 (c) lim f x does not exist. 53. g x 1 2 x 0 55. f x 8 x 10 2 Vertical asymptote at x 2x2 x x2 x sin 4x 5x 2x 1 x 2 1 1 Vertical asymptote at x x x3 1 1 1 x3 10 1 x 1 3 57. x→ 2 lim 59. x→ 1 lim x→ 1 lim x2 1 61. lim x→1 63. lim x x→0 65. lim x→0 x→0 lim 4 sin 4x 5 4x 4 5 67. lim x→0 csc 2x x x →0 lim 1 x sin 2x 69. C 80,000p , 0 ≤ 0 < 100 100 p $14,117.65 $720,000 (b) C 50 (d) p → 100 (a) C 15 (c) C 90 $80.000 80,000p 100 p lim Problem Solving for Chapter 1 1. (a) Perimeter PAO x2 x2 Perimeter PBO x x y x2 1 1 2 2 1 2 2 x2 x2 x2 x2 y2 x4 y2 x4 1 1 1 1 (b) r x x2 x x2 12 12 x4 4 x2 x2 2 9.08 9.60 0.95 x4 x4 1 1 1 0.1 2.10 2.00 1.05 0.01 2.01 2.00 1.005 1 y2 x4 x Perimeter Perimeter rx PAO PBO 33.02 33.77 0.98 3.41 3.41 1 (c) lim r x x→0 1 1 0 0 1 1 2 2 1 50 Chapter 1 Limits and Their Properties 12 5 5 . 12 3. (a) There are 6 triangles, each with a central angle of 60 3. Hence, Area hexagon 6 1 bh 2 6 1 1 sin 2 3 5. (a) Slope (b) Slope of tangent line is y h = sin θ 33 2 h = sin 60° 1 60° 2.598. 12 y 5 x 12 5 x 12 x, 169 x x→5 5 169 Tangent line 12 169 x2 5 x x x 12 169 5 x2 12 12 169 169 x2 x2 x2 1 θ (c) Q mx x, y Error: 33 2 0.5435. (d) lim mx x→5 lim 12 (b) There are n triangles, each with central angle of 2 n. Hence, An (c) n 1 bh 2 6 2.598 n 1 2 1 sin 2 n 24 3.106 48 3.133 n sin 2 2 96 3.139 n . x→5 lim lim 144 169 x2 5 12 169 x2 x2 5 12 x 5 169 5 12 25 169 x2 x2 x→5 n An 12 3 x→5 lim 12 12 10 (d) As n gets larger and larger, 2 Letting x An 2 n, sin 2 n 2n . sin x x n approaches 0. 12 This is the same slope as part (b). sin 2 n 2n which approaches 1 7. (a) 3 x1 x1 3 3 ≥0 ≥ (b) 3 27 27, x 3 x x x1 x2 1 3 − 30 0.5 (c) x → 27 lim f x 3 2 28 27 1 3 27 1 1 14 2 x≥ Domain: x ≥ (d) lim f x x→1 0.0714 12 − 0.1 1 x1 3 x→1 lim 2 3 1 x1 3 1 x2 x1 3 3 3 4 3 x1 x1 2 3 3 2 2 x→1 lim lim 3 1 x1 3 x1 3 x1 1 3 3 x→1 1 1 x1 3 3 2 x1 3 2 x→1 lim 1 2 3 1 12 1 9. (a) lim f x x→2 1 12 3: g1, g4 (b) f continuous at 2: g1 (c) lim f x x→2 3: g1, g3, g4 Problem Solving for Chapter 1 11. 4 3 2 1 −4 −3 −2 −1 −2 −3 −4 a b x x 1 2 3 4 1 2 y 51 13. (a) y (a) f1 f0 f f 1 2 1 0 0 3 1 1 1 1 2 1 1 1 0 (b) (i) lim Pa, b x x→a 1 0 0 1 (ii) lim Pa, b x x→a 1 1 (iii) lim Pa, b x x→b 2.7 x→1 (iv) lim Pa, b x x→b (b) lim f x lim f x (c) Pa, b is continuous for all positive real numbers except x a, b. (d) The area under the graph of u, and above the x-axis, is 1. x→1 x→1 2 lim f x (c) f is continuous for all real numbers except x 0, ± 1, ± 2, ± 3, . . . ...
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