EVNREV01 - 324 Chapter 1 Limits and Their Properties Review...

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Unformatted text preview: 324 Chapter 1 Limits and Their Properties Review Exercises for Chapter 1 2. Precalculus. L 4. 9 1 2 3 1 2 8.25 x fx lim f x 0.1 0.358 0.2 0.01 0.354 0.001 0.354 0.001 0.354 0.01 0.353 0.1 0.349 −1 0.5 1 x→0 − 0.5 6. g x 3x x 2 (a) lim g x does not exist. x→2 (b) lim g x x→0 0 8. lim Let x→9 x 9 3. Assuming 4 < x < 16, you can choose Hence, for 0 < x x x 3< 9< x x 3 3 x x fx 9< x 5 , you have 3 5. > 0 be given. We need x 3< ⇒ x 3 9 <5 < 3< L< 10. lim 9 x→5 9. Let > 0 be given. can be any positive 12. lim 3 y y→4 1 34 1 9 number. Hence, for 0 < x 9 fx t2 t 9 3 9< L< 5 < , you have 14. lim t→3 lim t t→3 3 6 16. lim 4 x x 2 x→0 x→0 lim 4 x 1 4 x 2 1 4 4 4 x x 2 2 x→0 lim x 2 18. lim 1 1 s s 1 s→0 s→0 lim 1 1 s s 1 1 1 1 1 s→0 1 1 1 s s s 1 1 1 1 1 s 1 1 2 s→0 lim 11 s s1 1 s lim 20. lim x→ 2 x2 x3 4 8 x→ 2 lim x x x x2 2x 2 2 x2 2x 4 2 2x 4 22. x→ lim 4 4x tan x 4 1 4 x→ 2 lim 4 12 1 3 Review Exercises for Chapter 1 cos x x 1 cos cos x cos x x 01 2 3 7 12 325 24. lim x→0 lim lim 0 3 4 x→0 sin x 1 sin x 1 sin x x x→0 lim sin x→0 0 26. lim f x x→c 2g x 2 28. f x (a) x 1 x 3 x 1 1.1 0.3228 1.01 0.3322 1.001 0.3332 1.0001 0.3333 1 3. 3 3 (b) −3 2 fx 1 x 1 x 3 3 x→1 lim x 1 0.333 lim lim 1 x 3 Actual limit is x 1 1 1 1 3 3 3 −3 3 (c) lim x→1 x 1 x→1 x x 3 x 2 x 2 x→1 x 1 11 1 3 x x x 2 x2 x→1 lim 1 3 x 3 30. s t 0⇒ 4.9t2 200 0 ⇒ t2 40.816 ⇒ t 6.39 sec When t lim t→a 6.39, the velocity is approximately sa a st t lim t→a t → 6.39 4.9 a t 6.39 62.6 m sec. lim 4.9 6.39 32. lim x x→4 1 does not exist. The graph jumps from 2 to 3 4. 34. lim g x x→1 1 1 2. at x 36. lim f s s→ 2 2 38. f x lim f x 3x2 0, x lim x 1 3x 2 2, x x 2 1 x x 2 5 1 1 x→1 x→1 x→1 lim 3x 0 1 Removable discontinuity at x Continuous on ,1 1, 326 Chapter 1 5 x, 2x 3, x 3 3 1 Limits and Their Properties x≤2 x>2 40. f x x→2 x→2 42. f x lim 1 x x 1 x , 1 1 1 x lim 5 2x x→0 lim Domain: 2 1 , 0, 0 Nonremovable discontinuity at x Continuous on ,2 2, x 2x x 12x 1 2 1 1 1 2 Nonremovable discontinuity at x Continuous on ,1 0, 46. f x tan 2x 44. f x lim Nonremovable discontinuities when x 1 1, 2n 4 2n 4 1 x→ Removable discontinuity at x Continuous on ,1 Continuous on 1 , 2n 4 1 for all integers n. 48. lim x x→1 x→3 1 1 2 4 x→1 lim x Find b and c so that lim x2 Consequently we get Solving simultaneously, 50. C 9.80 9.80 2.50 2.50 x x 1 1 b b bx c c 2 2 and lim x2 x→3 bx c c c 4. 4. 52. f x 4. and 9 3 and 3b 1, x > 0 x 1x ,0 0 0 1, (a) Domain: (b) lim f x x→0 C has a nonremovable discontinuity at each integer. 30 (c) lim f x x→1 0 0 5 54. h x 4x 4 x2 2 and x 2 56. f x csc x Vertical asymptotes at x Vertical asymptote at every integer k 58. x→ 1 2 lim x 2x x2 x 1 2x 1 1 60. x→ 1 lim x x4 1 x2 1 1 x→ 1 lim x2 1 1x 1 1 4 62. x→ 1 lim 64. lim x→2 3 4 66. lim x→0 sec x x 68. lim x→0 cos2 x x Problem Solving for Chapter 1 tan 2x x x fx lim tan 2x x 0.1 2.0271 2 0.01 2.0003 0.001 2.0000 0.001 2.0000 0.01 2.0003 0.1 2.0271 327 70. f x (a) x→0 (b) Yes, define fx tan 2x , x 2, x x 0 . 0 0. Now f x is continuous at x Problem Solving for Chapter 1 2. (a) Area Area (b) a x PAO PBO Area Area 1 bh 2 1 bh 2 PBO PAO 4 PAO PBO 2 8 4 lim x 2 1 2 2 0 1 1x 2 1 1y 2 x2 2 x2 1 12 12 1 x (c) Let Q x Area Area ax (c) lim a x x→0 x 2 y 2 x2 2 4. (a) Slope (b) Slope 4 3 3 4 0 0 4 3 Tangent line: y 4 y 3 x 4 3 x 4 3 25 4 x, y 25 x x→3 x, x2 3 4 25 x x x 25 25 3 3 3 3 25 x2 0.1 1 20 1 200 1 10 0.01 1 200 1 20,000 1 100 mx (d) lim mx x→3 lim x2 3 x2 25 4 16 x2 x x2 4 25 25 4 4 6 4 x2 x2 4 4 x→3 lim lim x→0 x→3 x3 25 x x2 4 x→3 lim 3 4 This is the slope of the tangent line at P. a bx x 3 a a a bx x bx bx 3 3 3 a a bx bx 3 3 6. 8. lim f x x→0 x→0 x→0 lim a2 ax tan x 2 a2 2 x→0 lim f x x Letting a Thus, x→0 x→0 lim a because lim tan x x 1 3 simplifies the numerator. Thus, a2 2 2 1 a a 0 0 1, 2 lim 3 bx x 3 x→0 lim x 3 3 bx bx b bx 6. 3 . a2 3 a a 2a x→0 lim Setting Thus, a b 3 3 3 and b 3, you obtain b 6. 328 10. Chapter 1 y 3 2 1 −1 Limits and Their Properties (a) f 1 4 4 f3 f1 1 3 4 0 1 (b) lim f x x→1 x→1 x→0 x→0 1 0 lim f x lim f x (c) f is continuous for all real numbers except x 0, ± 1, ± 1, ± 1, . . . 2 3 1 x −1 −2 1 lim f x 12. (a) v2 192,000 r r lim r 192,000 r v2 v02 v02 48 48 14. Let a 0 and let > 0 be given. There exists 1 > 0 L<. such that if 0 < x 0 < , then f x 0< Let 1 a . Then for 0 < x 1 a, you have x< ax < f ax 1 v 192,000 v02 48 a 1 v→0 192,000 48 v02 48 1920 r v2 v02 4 3 feet sec. v02 2.17 2.17 L<. 1 2 x x 0 . 0 Let v0 (b) v2 1920 r r lim r As a counterexample, let f x Then lim f x x→0 1 x→0 L, 2. but lim f ax x→0 lim f 0 v2 1920 v02 2.17 v→0 1920 2.17 v02 2.17 mi sec v2 10,600 v02 6.99 1.47 mi sec . Let v0 (c) r lim r v→0 10,600 6.99 v02 6.99 2.64 mi sec. Let v0 Since this is smaller than the escape velocity for earth, the mass is less. ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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