(Solution Manual) Calculus Early Transcendentals

Stewart, 8

Math & Statistics

Solution Manual

8th Edition

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Solution Manual for Calculus Early Transcendentals, The downloadable file consists of 302 pages, and the total size of the file is 7.89 MB. Purchase now only for highly discounted price.

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− 9 42 − 36 6 42 − 38 4 − Solutions Manual for Calculus Early Transce ndentals 8th Editio n by Stewart Ibsn 2 LIMITS AND DERIVATIVES 2.1 The Tangent and Velocity Problems 1. (a) Using P (15 250), we construc t the following table: t slope = mPQ 5 (5 694) 694−250 = − 444 = −44 4 5−15 10 10 (10 444) 444−250 = − 194 = −38 8 10−15 5 20 (20 111) 111−250 = − 139 = −27 8 20−15 5 25 (25 28) 28−250 = − 222 = −22 2 25−15 10 30 (30 0) 0−250 = − 250 = −16 6 30−15 15 (c) From the graph, we can estimate the slope of the tangen t line at P to be −300 = −33 3. (b) Using the values of t that correspond to the points closest to P (t = 10 and t = 20), we have −38 8+ (−27 8) = 33 3 2 2. (a) Slope = 2948 − 2530 = 418 ≈ 69 67 (b) Slope = 2948 − 2661 = 287 = 71 75 (c) Slope = 2948 − 2806 = 142 = 71 (d) Slope = 3080 − 2948 = 132 = 66 42 − 40 2 44 − 42 2 From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeats minute after 42 minutes. After being stable for a while, the patient’s heart rate is dropping. 3. (a) y = 1 , P (2 1) 1 − x (b) The slope appears to be 1. (c) Using m = 1, an equation of the tangent line to the curve at P (2 −1) is y − (−1) = 1(x − 2), or y = x − 3. x (x 1 (1 − x)) mPQ (i) 1 5 (1 5 −2) (1 9 −1 111 111) (1 99 −1 010 101) (1 999 −1 001 001) (2 5 −0 666 667) (2 1 −0 909 091) (201 −0990 099) 2 (ii) 1 9 1 111 111 (iii) 1 99 1 010 101 (iv) 1 999 1 001 001 (v) 2 5 0 666 667 (vi) 2 1 0 909 091 2 − − −24h − 16h − − 6 − − − 68 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 4. (a) y = cos x, P (0 5 0) (b) The slope appears to be − . (c) y − 0 = − (x − 0 5) or y = − x + 1 . (d) 5. (a) y = y(t) = 40t − 16t2. At t = 2, y = 40(2) − 16(2)2 = 16. The average velocity between times 2 and 2+ h is vave = y(2 + h) − y(2) = (2 + h) − 2 40(2 + h) 16(2 + h)2 16 = h 2 = 24 16h, if h = 0. h (i) [2 2 5]: h = 0 5, vave = −32 ft s (ii) [2 2 1]: h = 0 1, vave = −25 6 ft s (iii) [2 2 05]: h = 0 05, vave = −24 8 ft s (iv) [2 2 01]: h = 0 01, vave = −24 16 ft s (b) The instantaneous velocity when t = 2 (h approaches 0) is −24 ft s. 6. (a) y = y(t) = 10t − 1 86t2. At t = 1, y = 10(1) − 1 86(1)2 = 8 14. The average velocity between times 1 and 1+ h is vave = y(1 + h) − y(1) = (1 + h) − 1 10(1 + h) 1 86(1 + h)2 8 14 = h 6 28h 1 86h2 h = 6 28 − 1 86h, if h 6= 0. (i) [1 2]: h = 1, vave = 4 42 m s (ii) [1 1 5]: h = 0 5, vave = 5 35 m s (iii) [1 1 1]: h = 0 1, vave = 6 094 m s (iv) [1 1 01]: h = 0 01, vave = 6 2614 m s (v) [1 1 001]: h = 0 001, vave = 6 27814 m s (b) The instantaneo us velocit y when t = 1 (h approaches 0) is 6 28 m s. 7. (a) (i) On the interval [2 4] , v = s(4) − s(2) = 79 2 − 20 6 = 29 3 ft s. ave 4 − 2 2 (ii) On the interval [3 4] , v = s(4) − s(3) = 79 2 − 46 5 = 32 7 ft s. ave 4 − 3 1 (iii) On the interval [4 5] , v = s(5) − s(4) = 124 8 − 79 2 = 45 6 ft s. ave 5 − 4 1 (iv) On the interval [4 6] , v = s(6) − s(4) = 176 7 − 79 2 = 48 75 ft s. ave 6 − 4 2 x mPQ (i) 0 (0 1) (0 4 0 309017) (0 49 0 03141 1) (0 499 0 0031 42) (1 −1) (0 6 −0 309017) (0 51 −0 031411) (0 501 −0 003142) −2 −3 090170 −3 141076 −3 141587 −2 −3 090170 −3 141076 −3 141587 (ii) 0 4 (iii) 0 49 (iv) 0 499 (v) 1 (vi) 0 6 (vii) 0 51 (viii) 0 501 °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicl y accessibl e website , in whole or in part. SECTION 2.1 THE TANGENT AND VELOCITY PROBLEMS ¤ 69 (b) Using the points (2 16) and (5 105) from the approximate tangent line, the instantaneous velocity at t = 3 is about 105 − 16 = 89 ≈ 29 7 ft s. 5 − 2 3 8. (a) (i) s = s(t) = 2 sin t +3 cos t. On the interval [1 2], v = s(2) − s(1) = 3 − (−3) = 6 cm s. ave 2 − 1 1 (ii) On the interval [1 1 1], v = s(1 1) − s(1) ≈ −3 471 − (−3) = −4 71 cm s. ave 1 1 − 1 0 1 (iii) On the interval [1 1 01], v = s(1 01) − s(1) ≈ −3 0613 − (−3) = −6 13 cm s. ave 1 01 − 1 0 01 (iv) On the interval [1 1 001], v = s(1 001) − s(1) ≈ −3 00627 − (−3) = −6 27 cm s. ave 1 001 − 1 0 001 (b) The instantaneo us velocit y of the particl e when t = 1 appears to be about −6 3 cm s. 9. (a) For the curve y = sin(10 x) and the point P (1 0): As x approaches 1, the slopes do not appear to be approaching any particular value. (b) We see that problems with estimation are caused by the frequent oscillations of the graph. The tangent is so steep at P that we need to take x-values much closer to 1 in order to get accurate estimates of its slope. (c) If we choose x = 1 001, then the point is (1 001 −0 0314) and mPQ ≈ −31 3794. If x = 0 999, then is (0 999 0 0314) and mPQ = −31 4422. The average of these slopes is −31 4108. So we estimat e that the slope of the tangen t line at P is abou t −31 4. x mPQ 2 (2 0) 0 1 5 (1 5 0 8660) 1 7321 1 4 (1 4 −0 4339) −1 0847 1 3 (1 3 −0 8230) −2 7433 1 2 (1 2 0 8660) 4 3301 1 1 (1 1 −0 2817) −2 8173 x mPQ 0 5 (0 5 0) 0 0 6 (0 6 0 8660) −2 1651 0 7 (0 7 0 7818) −2 6061 0 8 (0 8 1) −5 0 9 (0 9 −0 3420) 3 4202 ° — − − +x − 70 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 2.2 The Limit of a Function 1. As x approaches 2, ƒ(x) approaches 5. [Or, the values of ƒ(x) can be made as close to 5 as we like by taking x sufficiently close to 2 (but x 6= 2).] Yes, the graph could have a hole at (2 5) and be defined such that ƒ (2) = 3. 2. As x approaches 1 from the left, ƒ(x) approaches 3; and as x approaches 1 from the right, ƒ(x) approaches 7. No, the limit does not exist because the left- and right -hand limits are different. 3. (a) lim x→− 3 ƒ(x) = ∞ means that the values of ƒ(x) can be made arbitrarily large (as large as we please) by taking x sufficiently close to −3 (but not equal to −3). (b) lim x→4+ ƒ(x) = −∞ means that the values of ƒ(x) can be made arbitrarily large negative by taking x sufficiently close to 4 through values larger than 4. 4. (a) As x approaches 2 from the left, the values of ƒ(x) approach 3, so lim x→2− (b) As x approaches 2 from the right, the values of ƒ(x) approach 1, so lim ƒ(x) = 3. ƒ(x) = 1. x→2+ (c) lim ƒ(x) does not exist since the left-hand limit does not equal the right -hand limit. x→2 (d) When x = 2, y = 3, so ƒ (2) = 3. (e) As x approaches 4, the values of ƒ(x) approach 4, so lim ƒ(x) = 4. x→4 (f ) There is no value of ƒ(x) when x = 4, so ƒ (4) does not exist. 5. (a) As x approaches 1, the values of ƒ(x) approach 2, so lim ƒ(x) = 2. x→1 (b) As x approaches 3 from the left, the values of ƒ(x) approach 1, so lim x→3− (c) As x approaches 3 from the right, the values of ƒ(x) approach 4, so lim ƒ(x) = 1. ƒ(x) = 4. x→3+ (d) lim ƒ(x) does not exist since the left-hand limit does not equal the right -hand limit. x→3 (e) When x = 3, y = 3, so ƒ (3) = 3. 6. (a) h(x) approaches 4 as x approaches 3 from the left, so lim h(x) = 4. x→− 3 (b) h(x) approaches 4 as x approaches 3 from the right, so lim h(x) = 4. →− 3 (c) lim x→− 3 h(x) = 4 because the limits in part (a) and part (b) are equal. (d) h(−3) is not defined, so it doesn’t exist. (e) h(x) approaches 1 as x approaches 0 from the left, so lim x→0− h(x) = 1. (f ) h(x) approaches 1 as x approaches 0 from the right, so lim x→0+ h(x) = −1. (g) lim h(x) does not exist because the limits in part (e) and part (f ) are not equal. x→0 (h) h(0) = 1 since the point (0 1) is on the graph of h. (i) Since lim x→2− h(x) = 2 and lim x→2+ h(x) = 2, we have lim h(x) = 2. x → 2 (j) h(2) is not defined, so it doesn’t exist. °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicl y accessibl e website , in whole or in part. ∞ −∞ −∞ ∞ ∞ − 2 x if x 1 sin x if x SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 71 (k) h(x) approaches 3 as x approaches 5 from the right, so lim x→5+ h(x) = 3. (l) h(x) does not approach any one number as x approaches 5 from the left, so lim x→5− h(x) does not exist. 7. (a) lim g(t) = −1 (b) lim g(t) = −2 t→0− t→0+ (c) lim g(t) does not exist because the limits in part (a) and part (b) are not equal. t→0 (d) lim g(t) = 2 (e) lim g(t) = 0 t→2− t→2+ (f ) lim g(t) does not exist because the limits in part (d) and part (e) are not equal. t→2 (g) g(2) = 1 (h) lim g(t) = 3 t→4 8. (a) lim A(x) = ∞ (b) lim A(x) does not exist. (c) lim A(x) = −∞ x→− 3 (d) lim x→2+ x→2 A(x) = (e) lim x→− 1 A(x) = −∞ x→2− (f ) The equations of the vertical asymptotes are x = −3, x = −1 and x = 2. 9. (a) lim x→− 7 (d) lim x→6− ƒ(x) = (b) lim x→− 3 ƒ(x) = (e) lim x→6+ ƒ(x) = (c) lim ƒ(x) = x→0 ƒ(x) = ∞ (f ) The equations of the vertical asymptotes are x = −7, x = −3, x = 0, and x = 6. 10. lim t→12− ƒ(t) = 150 mg and lim t→12+ ƒ(t) = 300 mg. These limits show that there is an abrupt change in the amount of drug in the patient’s bloodstream at t = 12 h. The left -hand limit represents the amount of the drug just before the fourth injection. The right -hand limit represents the amount of the drug just after the fourth injection. 11. From the graph of 1+ x if x −1 ƒ(x) = x2 if −1 ≤ x 1 , — ≥ we see that lim ƒ(x) exists for all a except a = 1. Notice that the x→ right and left limits are different at a = −1. 12. From the graph of 1+ sin x if x 0 ƒ(x) = cos x if 0 ≤ x ≤ , we see that lim ƒ(x) exists for all a except a = . Notice that the x→ right and left limits are different at a = . °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicly accessibl e website , in whole or in part. − 72 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 13. (a) lim x→0− (b) lim x→0+ ƒ(x) = 1 ƒ(x) = 0 (c) lim ƒ(x) does not exist because the limits x→0 in part (a) and part (b) are not equal. 14. (a) lim x→0− (b) lim x→0+ ƒ(x) = −1 ƒ(x) = 1 (c) lim ƒ(x) does not exist because the limits x→0 in part (a) and part (b) are not equal. 15. lim ƒ(x) = −1, lim ƒ(x) = 2, ƒ (0) = 1 16. lim ƒ(x) = 1, lim ƒ(x) = −2, lim ƒ(x) = 2, x→0− x→0+ x→0 x→3− x→3+ ƒ (0) = −1, ƒ (3) = 1 17. lim x→3+ ƒ(x) = 4, lim x→3− ƒ(x) = 2, lim x→− 2 ƒ(x) = 2, 18. lim x→0− ƒ(x) = 2, lim x→0+ ƒ(x) = 0, lim x→4− ƒ(x) = 3, ƒ (3) = 3, ƒ(−2) = 1 lim x→4+ ƒ(x) = 0, ƒ (0) = 2, ƒ (4) = 1 19. For ƒ(x) = x2 3x : x2 − 9 It appears that lim x→3 x2 − 3x x2 − 9 = 1 . 2 x ƒ(x) 3 1 0 508 197 3 05 0 504 132 3 01 0 500 832 3 001 0 500 083 3 0001 0 500 008 x ƒ(x) 2 9 0 491 525 2 95 0 495 798 2 99 0 499 165 2 999 0 499 917 2 9999 0 499 992 °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicl y accessibl e website , in whole or in part. © Cengage Learning. All Rights Reserved. →− 3 − − − − 20. For ƒ(x) = x2 3x : x2 − 9 SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 73 It appears that x lim + ƒ(x) = −∞ and that lim ƒ(x) = ∞, so lim x2 3x 2 does not exist. x→− 3− x→−3 x − 9 21. For ƒ(t) = e5t 1 : t 22. For ƒ(h) = (2 + h)5 32 : h It appears that lim t→0 e5t − 1 t = 5. It appears that lim h→0 (2 + h)5 − 32 h = 80. 23. For ƒ(x) = ln x − ln 4 : x − 4 It appear s that lim ƒ(x) = 0 25. The graph confirms that result. x→4 1+ p9 24. For ƒ(p) = : 1+ p15 It appears that lim p→− 1 ƒ(p) = 0 6. The graph confirms that result. t ƒ(t) 0 5 22 364 988 0 1 6 487 213 0 01 5 127 110 0 001 5 012 521 0 0001 5 001 250 t ƒ(t) −0 5 1 835 830 −0 1 3 934 693 −0 01 4 877 058 −0 001 4 987 521 −0 0001 4 998 750 h ƒ(h) 0 5 131 312 500 0 1 88 410 100 0 01 80 804 010 0 001 80 080 040 0 0001 80 008 000 h ƒ(h) −0 5 48 812 500 −0 1 72 390 100 −0 01 79 203 990 −0 001 79 920 040 −0 0001 79 992 000 p ƒ(p) −1 1 0 427 397 −1 01 0 582 008 −1 001 0 598 200 −1 0001 0 599 820 p ƒ(p) −0 9 0 771 405 −0 99 0 617 992 −0 999 0 601 800 −0 9999 0 600 180 x ƒ(x) −2 5 −5 −2 9 −29 −2 95 −59 −2 99 −299 −2 999 −2999 −2 9999 −29,999 x ƒ(x) −3 5 7 −3 1 31 −3 05 61 −3 01 301 −3 001 −3 0001 3001 30,001 x ƒ(x) 3 9 0 253 178 3 99 0 250 313 3 999 0 250 031 3 9999 0 250 003 x ƒ(x) 4 1 0 246 926 4 01 0 249 688 4 001 0 249 969 4 0001 0 249 997 © Cengage Learning. All Rights Reserved. °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicly accessibl e website , in whole or in part. © Cengage Learning. All Rights Reserved. ≈ − 74 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 25. For ƒ( ) = sin 3 : tan 2 It appears that lim sin 3 = 1 5. →0 tan 2 The graph confirms that result. 26. For ƒ(t) = 5t 1 : t It appear s that lim ƒ(t) 1 6094. The graph confirms that result. t→0 27. For ƒ(x) = xx: It appears that lim x→0+ ƒ(x) = 1. The graph confirms that result. 28. For ƒ(x) = x2 ln x: It appears that lim x→0+ ƒ(x) = 0. The graph confirms that result. 29. (a) From the graphs , it seems that lim cos 2x − cos x = −1 5. (b) x→0 x2 t ƒ(t) 0 1 1 746 189 0 01 1 622 459 0 001 1 610 734 0 0001 1 609 567 t ƒ(t) −0 1 −0 01 −0 001 −0 0001 1 486 601 1 596 556 1 608 143 1 609 308 ƒ( ) ±0 1 1 457 847 ±0 01 1 499 575 ±0 001 1 499 996 ±0 0001 1 500 000 x ƒ(x) 0 1 0 794 328 0 01 0 954 993 0 001 0 993 116 0 0001 0 999 079 x ƒ(x) 0 1 0 01 0 001 0 0001 −0 023 026 −0 000 461 −0 000 007 −0 000 000 x ƒ(x) ±0 1 −1 493 759 ±0 01 −1 499 938 ±0 001 −1 499 999 ±0 0001 −1 500 000 © Cengage Learning. All Rights Reserved. °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicl y accessibl e website , in whole or in part. © Cengage Learning. All Rights Reserved. ∞ → — → → — −∞ ∞ 30. (a) From the graphs, it seems that lim sin x SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 75 = 0 32. (b) x→0 sin x 31. lim Later we will be able to show that the exact value is 1 . x +1 = ∞ since the numerator is positive and the denominator approaches 0 from the positive side as x → 5+. x→5+ x − 5 32. lim x +1 = −∞ since the numerator is positive and the denominator approaches 0 from the negative side as x → 5−. x→5− x − 5 33. lim 2 − x = since the numerator is positive and the denominator approaches 0 through positive values as x 1. x→1 (x − 1)2 √x 34. lim x→3− (x − 3)5 = −∞ since the numerator is positive and the denominator approaches 0 from the negative side as x → 3−. 35. Let t = x2 9. Then as x 3+, t 0+, and lim x→3+ ln(x2 9) = lim ln t = by (5). t→0+ 36. lim x→0+ ln(sin x) = −∞ since sin x → 0+ as x → 0+. 37. lim 1 sec x = −∞ since 1 is posit ive and sec x → −∞ as x → ( 2)+. x→( 2)+ x x 38. lim cot x = lim cos x = −∞ since the numerator is negative and the denominator approaches 0 through positive values x→ − as x → −. x→ − sin x 39. lim x csc x = lim x = −∞ since the numerator is positive and the denominator approaches 0 through negative x→2 − x→2 − sin x values as x → 2 −. x2 − 2x x(x − 2) x 40. lim x→2− 2 = lim x − 4x +4 x→2− 2 = lim (x − 2) x→2− x − 2 = −∞ since the numerator is positive and the denominator approaches 0 through negative values as x → 2−. 41. lim x→2+ x2 − 2x − 8 — x→2+ (x − 4)(x + 2) = since the numerator is negative and the denominator approaches 0 through (x − 3)(x − 2) x ƒ(x) ±0 1 0 323 068 ±0 01 0 318 357 ±0 001 ±0 0001 0 318 310 0 318 310 © Cengage Learning. All Rights Reserved. x2 5x +6 = lim x→0 x x negative values as x → 2+. 42. lim 1 − ln x = ∞ since 1 →∞ and ln x → −∞ as x → 0+. 43. lim (ln x2 − x−2) = −∞ since ln x2 → −∞ and x−2 →∞ as x → 0. °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicly accessibl e website , in whole or in part. x→0+ © Cengage Learning. All Rights Reserved. 2 −∞ −∞ ≈ 76 ¤ CHAPTER 2 LIMITS AND DERIVATIVES x2 +1 x2 +1 44. (a) The denominator of y = 3x − 2x2 = x(3 − 2x) is equal to zero when (b) x = 0 and x = 3 (and the numerato r is not), so x = 0 and x = 1 5 are vertical asymptotes of the function. 45. (a) ƒ(x) = 1 . x3 − 1 From these calculations, it seems that lim x→1− ƒ(x) = and lim x→1+ ƒ(x) = ∞. (b) If x is slightly smaller than 1, then x3 − 1 will be a negative number close to 0, and the reciprocal of x3 − 1, that is, ƒ(x), will be a negative number with large absolute value. So lim x→1− ƒ(x) = −∞. If x is slightly larger than 1, then x3 − 1 will be a small positive number, and its reciprocal, ƒ(x), will be a large positive number. So lim x→1+ ƒ(x) = ∞. (c) It appears from the graph of ƒ that lim x→1− ƒ(x) = and lim x→1+ ƒ(x) = ∞. 46. (a) From the graphs, it seems that lim tan 4x = 4. (b) x→0 x 47. (a) Let h(x) = (1 + x)1 x. (b) It appears that lim (1 + x)1 x 2 71828, which is approximatel y e. x→0 x ƒ(x) 0 5 0 9 0 99 0 999 0 9999 0 99999 −1 14 −3 69 −33 7 −333 7 −3333 7 −33,333 7 x ƒ(x) 1 5 0 42 1 1 3 02 1 01 33 0 1 001 333 0 1 0001 3333 0 1 00001 33,333 3 x ƒ(x) ±0 1 ±0 01 ±0 001 ±0 0001 4 227 932 4 002 135 4 000 021 4 000 000 x h(x) −0 001 −0 0001 −0 00001 −0 000001 0 000001 0 00001 0 0001 0 001 2 71964 2 71842 2 71830 2 71828 2 71828 2 71827 2 71815 2 71692 © Cengage Learning. All Rights Reserved. INSTRUCTOR USE ONLY In Section 3.6 we will see that the value of the limit is exactly e. °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicl y accessibl e website , in whole or in part. © Cengage Learning. All Rights Reserved. − SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 77 48. (a) No, because the calculator -produced graph of ƒ(x) = ex + ln |x − 4| looks like an exponential function, but the graph of ƒ has an infinite discontinuity at x = 4. A second graph, obtained by increasing the numpoints option in Maple, begins to reveal the discontinuity at x = 4. (b) There isn’t a single graph that shows all the features of ƒ. Several graphs are needed since ƒ looks like ln |x − 4| for large negative values of x and like ex for x 5, but yet has the infinite discontiuity at x = 4. A hand -drawn graph, though distorted, might be better at revealing the main features of this function. 49. For ƒ(x) = x2 − (2x 1000 ): (a) It appears that lim ƒ(x) = 0. x→0 50. For h(x) = tan x − x: x3 (a) (b) It appear s that lim ƒ(x) = 0 001. x→0 (b) It seems that lim h(x) = 1 . x→0 3 x ƒ(x) 0 04 0 02 0 01 0 005 0 003 0 001 0 000 572 −0 000 614 −0 000 907 −0 000 978 −0 000 993 −0 001 000 x ƒ(x) 1 0 998 000 0 8 0 638 259 0 6 0 358 484 0 4 0 158 680 0 2 0 038 851 0 1 0 008 928 0 05 0 001 465 x h(x) 1 0 0 557 407 73 0 5 0 370 419 92 0 1 0 334 672 09 0 05 0 333 667 00 0 01 0 333 346 67 0 005 0 333 336 67 © Cengage Learning. All Rights Reserved. °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicly accessibl e website , in whole or in part. NOT FOR SALE © Cengage Learning. All Rights Reserved. INSTRUCTOR USE ONLY 78 ¤ CHAPTER 2 LIMITS AND DERIVATIVES (c) Here the values will vary from one calculator to another. Every calculator will eventually give false values . (d) As in part (c), when we take a small enough viewing rectangle we get incorrect output. 51. No matter how many times we zoom in toward the origin, the graph s of ƒ(x) = sin( x) appear to consis t of almos t-vertical lines. This indicates more and more frequent oscillations as x → 0. 52. (a) For any positive integer n, if x = 1 , then ƒ(x) = tan 1 = tan(n ) = 0. (Remember that the tangent function has period .) n x °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicl y accessibl e website , in whole or in part. x h(x) 0 001 0 333 333 50 0 0005 0 333 333 44 0 0001 0 333 330 00 0 00005 0 333 336 00 0 00001 0 333 000 00 0 000001 0 000 000 00 NOT FOR SALE © Cengage Learning. All Rights Reserved. 2 4 − x 4 4 4 4 4 4 4 reference angle for sin−1 . So x = ± − sin−1 are also equations of 3 (b) For any nonnegative number n, if x = 4 (4n + 1) , then SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 79 ƒ(x) = tan 1 = tan (4n + 1) = tan 4n + = tan n + = tan = 1 (c) From part (a), ƒ(x) = 0 infinitely often as x → 0. From part (b), ƒ(x) = 1 infinitely often as x → 0. Thus, lim tan 1 does not exist since ƒ(x) does not get close to a fixed number as x → 0. x→0 x 53. There appear to be vertical asymptotes of the curve y = tan(2 sin x) at x ≈ ±0 90 and x ≈ ±2 24. To find the exact equations of these asymptotes, we note that the graph of the tangent function has vertical asymptotes at x = + n. Thus, we must have 2 sin x = + n, or equivalently, sin x = + n. Since 2 4 2 −1 ≤ sin x ≤ 1, we must have sin x = ± and so x = ± sin−1 (corresponding to x ≈ ±0 90). Just as 150◦ is the reference angle for 30◦, − sin−1 is the 54. lim m0 m = lim 4 4 vertical asymptote s (correspon ding to x ≈ ±2 24). . As v → c−, 1 − v2 c2 → 0+, and m → ∞. v→c− v→c− 1 − v2 c2 x3 − 1 55. (a) Let y = √x − 1 . From the table and the graph, we guess that the limit of y as x approaches 1 is 6. x3 1 (b) We need to have 5 5 √x − 1 6 5. From the graph we obtain the approximat e point s of intersectio n P (0 9314 5 5) and (1 0649 6 5). Now 1 − 0 9314 = 0 0686 and 1 0649 − 1 = 0 0649, so by requiring that x be withi n 0 0649 of 1, we ensur e that y is within 0 5 of 6. 2.3 Calculating Limits Using the Limit Laws 1. (a) lim [ƒ(x)+ 5g(x)] = lim ƒ(x) + lim [5g(x)] [Limit Law 1] = lim ƒ(x)+5 lim g(x) [Limit Law 3] (b) lim [g(x)]3 = lim g(x) 3 [Limi t Law 6] = ( −2) = −8 x→2 x→2 = 4 + 5(−2) = −6 x→2 x→2 x→2 x→2 x→2 x y 0 99 5 925 31 0 999 5 992 50 0 9999 5 999 25 1 01 6 075 31 1 001 6 007 50 1 0001 6 000 75 NOT FOR SALE © Cengage Learning. All Rights Reserved. INSTRUCTOR USE ONLY °c 2016 Cengage Learnin g. All Rights Reser ved. May not be scanned, copied , or duplic ated, or posted to a publicly accessibl e website , in whole or in part.