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Solution Manual for Calculus, 10th Edition

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    Limits and Continuity Exercise Set 1.1 1. (a) 3 (b) 3 (c) 3 (d) 3 2. (a) 0 (b) 0 (c) 0 (d) 0 3. (a) − 1 (b) 3 (c) does not exist (d) 1 4. (a) 2 (b) 0 (c) does not exist (d) 2 5. (a) 0 (b) 0 (c) 0 (d) 3 6. (a) 1 (b) 1 (c) 1 (d) 0 7. (a) −∞ (b) −∞ (c) −∞ (d) 1 8. (a) + ∞ (b) + ∞ (c) + ∞ (d) can not be found from graph 9. (a) + ∞ (b) + ∞ (c) 2 (d) 2 (e) −∞ (f ) x = − 2, x = 0, x = 2 10. (a) does not exist (b) −∞ (c) 0 (d) − 1 (e) + ∞ (f ) 3 (g) x = − 2, x = 2 11. (i) − 0 . 01 − 0 . 001 − 0 . 0001 0 . 0001 0 . 001 0 . 01 0 . 9950166 0 . 9995002 0 . 9999500 1 . 0000500 1 . 0005002 1 . 0050167 (ii) 0.995 1.005 -0.01 0.01 The limit appears to be 1. 12. (i) − 0 . 1 − 0 . 01 − 0 . 001 0 . 001 0 . 01 0 . 1 2 . 0135792 2 . 0001334 2 . 0000013 2 . 0000013 2 . 0001334 2 . 0135792 (ii) 2.014 2 1 . 0 1 . 0 - The limit appears to be 2. 39

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    40 Chapter 1 13. (a) 2 1 . 5 1 . 1 1 . 01 1 . 001 0 0 . 5 0 . 9 0 . 99 0 . 999 0 . 1429 0 . 2105 0 . 3021 0 . 3300 0 . 3330 1 . 0000 0 . 5714 0 . 3690 0 . 3367 0 . 3337 1 0 0 2 The limit is 1 / 3. (b) 2 1 . 5 1 . 1 1 . 01 1 . 001 1 . 0001 0 . 4286 1 . 0526 6 . 344 66 . 33 666 . 3 6666 . 3 50 0 1 2 The limit is + ∞ . (c) 0 0 . 5 0 . 9 0 . 99 0 . 999 0 . 9999 − 1 − 1 . 7143 − 7 . 0111 − 67 . 001 − 667 . 0 − 6667 . 0 0 -50 0 1 The limit is −∞ . 14. (a) − 0 . 25 − 0 . 1 − 0 . 001 − 0 . 0001 0 . 0001 0 . 001 0 . 1 0 . 25 0 . 5359 0 . 5132 0 . 5001 0 . 5000 0 . 5000 0 . 4999 0 . 4881 0 . 4721 0.6 0 -0.25 0.25 The limit is 1/2. (b) 0 . 25 0 . 1 0 . 001 0 . 0001 8 . 4721 20 . 488 2000 . 5 20001

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    Exercise Set 1.1 41 100 0 0 0.25 The limit is + ∞ . (c) − 0 . 25 − 0 . 1 − 0 . 001 − 0 . 0001 − 7 . 4641 − 19 . 487 − 1999 . 5 − 20000 0 -100 -0.25 0 The limit is −∞ . 15. (a) − 0 . 25 − 0 . 1 − 0 . 001 − 0 . 0001 0 . 0001 0 . 001 0 . 1 0 . 25 2 . 7266 2 . 9552 3 . 0000 3 . 0000 3 . 0000 3 . 0000 2 . 9552 2 . 7266 3 2 -0.25 0.25 The limit is 3. (b) 0 − 0 . 5 − 0 . 9 − 0 . 99 − 0 . 999 − 1 . 5 − 1 . 1 − 1 . 01 − 1 . 001 1 1 . 7552 6 . 2161 54 . 87 541 . 1 − 0 . 1415 − 4 . 536 − 53 . 19 − 539 . 5 60 -60 -1.5 0 The limit does not exist. 16. (a) 0 − 0 . 5 − 0 . 9 − 0 . 99 − 0 . 999 − 1 . 5 − 1 . 1 − 1 . 01 − 1 . 001 1 . 5574 1 . 0926 1 . 0033 1 . 0000 1 . 0000 1 . 0926 1 . 0033 1 . 0000 1 . 0000 1.5 1 -1.5 0 The limit is 1.

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    42 Chapter 1 (b) − 0 . 25 − 0 . 1 − 0 . 001 − 0 . 0001 0 . 0001 0 . 001 0 . 1 0 . 25 1 . 9794 2 . 4132 2 . 5000 2 . 5000 2 . 5000 2 . 5000 2 . 4132 1 . 9794 2.5 2 -0.25 0.25 The limit is 5/2. 17. False; define f ( x ) = x for x 6 = a and f ( a ) = a + 1. Then lim x → a f ( x ) = a 6 = f ( a ) = a + 1. 18. True; by 1.1.3. 19. False; define f ( x ) = 0 for x < 0 and f ( x ) = x + 1 for x ≥ 0. Then the left and right limits exist but are unequal. 20. False; define f ( x ) = 1 /x for x > 0 and f (0) = 2. 27. m sec = x 2 − 1 x + 1 = x − 1 which gets close to − 2 as x gets close to − 1, thus y − 1 = − 2( x + 1) or y = − 2 x − 1. 28. m sec = x 2 x = x which gets close to 0 as x gets close to 0, thus y = 0. 29. m sec = x 4 − 1 x − 1 = x 3 + x 2 + x + 1 which gets close to 4 as x gets close to 1, thus y − 1 = 4( x − 1) or y = 4 x − 3. 30. m sec = x 4 − 1 x + 1 = x 3 − x 2 + x − 1 which gets close to − 4 as x gets close to − 1, thus y − 1 = − 4( x +1) or y = − 4 x − 3. 31. (a) The length of the rod while at rest. (b) The limit is zero. The length of the rod approaches zero as its speed approaches c . 32. (a) The mass of the object while at rest. (b) The limiting mass as the velocity approaches the speed of light; the mass is unbounded. 33. (a) 3.5 2.5 –1 1 The limit appears to be 3. (b) 3.5 2.5 –0.001 0.001 The limit appears to be 3.

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    Exercise Set 1.2 43 (c) 3.5 2.5 –0.000001 0.000001 The limit does not exist. Exercise Set 1.2 1. (a) By Theorem 1.2.2, this limit is 2 + 2 · ( − 4) = − 6. (b) By Theorem 1.2.2, this limit is 0 − 3 · ( − 4) + 1 = 13. (c) By Theorem 1.2.2, this limit is 2 · ( − 4) = − 8. (d) By Theorem 1.2.2, this limit is ( − 4) 2 = 16. (e) By Theorem 1.2.2, this limit is 3 √ 6 + 2 = 2. (f ) By Theorem 1.2.2, this limit is 2 ( − 4) = − 1 2 . 2. (a) By Theorem 1.2.2, this limit is 0 + 0 = 0. (b) The limit doesn’t exist because lim f doesn’t exist and lim g does. (c) By Theorem 1.2.2, this limit is − 2 + 2 = 0. (d) By Theorem 1.2.2, this limit is 1 + 2 = 3. (e) By Theorem 1.2.2, this limit is 0 / (1 + 0) = 0. (f ) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. (g) The limit doesn’t exist because √ f ( x ) is not defined for 0 < x < 2. (h) By Theorem 1.2.2, this limit is √ 1 = 1. 3. By Theorem 1.2.3, this limit is 2 · 1 · 3 = 6. 4. By Theorem 1.2.3, this limit is 3 3 − 3 · 3 2 + 9 · 3 = 27. 5. By Theorem 1.2.4, this limit is (3 2 − 2 · 3) / (3 + 1) = 3 / 4. 6. By Theorem 1.2.4, this limit is (6 · 0 − 9) / (0 3 − 12 · 0 + 3) = − 3. 7. After simplification, x 4 − 1 x − 1 = x 3 + x 2 + x + 1, and the limit is 1 3 + 1 2 + 1 + 1 = 4. 8. After simplification, t 3 + 8 t + 2 = t 2 − 2 t + 4, and the limit is ( − 2) 2 − 2 · ( − 2) + 4 = 12. 9. After simplification, x 2 + 6 x + 5 x 2 − 3 x − 4 = x + 5 x − 4 , and the limit is ( − 1 + 5) / ( − 1 − 4) = − 4 / 5.

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    44 Chapter 1 10. After simplification, x 2 − 4 x + 4 x 2 + x − 6 = x − 2 x + 3 , and the limit is (2 − 2) / (2 + 3) = 0. 11. After simplification, 2 x 2 + x − 1 x + 1 = 2 x − 1, and the limit is 2 · ( − 1) − 1 = − 3. 12. After simplification, 3 x 2 − x − 2 2 x 2 + x − 3 = 3 x + 2 2 x + 3 , and the limit is (3 · 1 + 2) / (2 · 1 + 3) = 1. 13. After simplification, t 3 + 3 t 2 − 12 t + 4 t 3 − 4 t = t 2 + 5 t − 2 t 2 + 2 t , and the limit is (2 2 + 5 · 2 − 2) / (2 2 + 2 · 2) = 3 / 2. 14. After simplification, t 3 + t 2 − 5 t + 3 t 3 − 3 t + 2 = t + 3 t + 2 , and the limit is (1 + 3) / (1 + 2) = 4 / 3. 15. The limit is + ∞ . 16. The limit is −∞ . 17. The limit does not exist. 18. The limit is + ∞ . 19. The limit is −∞ . 20. The limit does not exist. 21. The limit is + ∞ . 22. The limit is −∞ . 23. The limit does not exist. 24. The limit is −∞ . 25. The limit is + ∞ . 26. The limit does not exist. 27. The limit is + ∞ . 28. The limit is + ∞ . 29. After simplification, x − 9 √ x − 3 = √ x + 3, and the limit is √ 9 + 3 = 6. 30. After simplification, 4 − y 2 − √ y = 2 + √ y , and the limit is 2 + √ 4 = 4. 31. (a) 2 (b) 2 (c) 2 32. (a) does not exist (b) 1 (c) 4 33. True, by Theorem 1.2.2. 34. False; e.g. lim x → 0 x 2 x = 0.

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    Exercise Set 1.2 45 35. False; e.g. f ( x ) = 2 x, g ( x ) = x , so lim x → 0 f ( x ) = lim x → 0 g ( x ) = 0, but lim x → 0 f ( x ) /g ( x ) = 2. 36. True, by Theorem 1.2.4. 37. After simplification, √ x + 4 − 2 x = 1 √ x + 4 + 2 , and the limit is 1 / 4. 38. After simplification, √ x 2 + 4 − 2 x = x √ x 2 + 4 + 2 , and the limit is 0. 39. (a) After simplification, x 3 − 1 x − 1 = x 2 + x + 1, and the limit is 3. (b) y x 4 1 40. (a) After simplification, x 2 − 9 x + 3 = x − 3, and the limit is − 6, so we need that k = − 6. (b) On its domain (all real numbers), f ( x ) = x − 3. 41. (a) Theorem 1.2.2 doesn’t apply; moreover one cannot subtract infinities. (b) lim x → 0 + ( 1 x − 1 x 2 ) = lim x → 0 + ( x − 1 x 2 ) = −∞ . 42. (a) Theorem 1.2.2 assumes that L 1 and L 2 are real numbers, not infinities. It is in general not true that ” ∞· 0 = 0 ”. (b) 1 x − 2 x 2 + 2 x = x 2 x ( x 2 + 2 x ) = 1 x + 2 for x 6 = 0, so that lim x → 0 ( 1 x − 2 x 2 + 2 x ) = 1 2 . 43. For x 6 = 1, 1 x − 1 − a x 2 − 1 = x + 1 − a x 2 − 1 and for this to have a limit it is necessary that lim x → 1 ( x + 1 − a ) = 0, i.e. a = 2. For this value, 1 x − 1 − 2 x 2 − 1 = x + 1 − 2 x 2 − 1 = x − 1 x 2 − 1 = 1 x + 1 and lim x → 1 1 x + 1 = 1 2 . 44. (a) For small x , 1 /x 2 is much bigger than ± 1 /x . (b) 1 x + 1 x 2 = x + 1 x 2 . Since the numerator has limit 1 and x 2 tends to zero from the right, the limit is + ∞ . 45. The left and/or right limits could be plus or minus infinity; or the limit could exist, or equal any preassigned real number. For example, let q ( x ) = x − x 0 and let p ( x ) = a ( x − x 0 ) n where n takes on the values 0 , 1 , 2. 46. If on the contrary lim x → a g ( x ) did exist then by Theorem 1.2.2 so would lim x → a [ f ( x ) + g ( x )], and that would be a contradiction. 47. Clearly, g ( x ) = [ f ( x ) + g ( x )] − f ( x ). By Theorem 1.2.2, lim x → a [ f ( x ) + g ( x )] − lim x → a f ( x ) = lim x → a [ f ( x ) + g ( x ) − f ( x )] = lim x → a g ( x ).

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    46 Chapter 1 48. By Theorem 1.2.2, lim x → a f ( x ) = ( lim x → a f ( x ) g ( x ) ) lim x → a g ( x ) = ( lim x → a f ( x ) g ( x ) ) · 0 = 0, since lim x → a f ( x ) g ( x ) exists. Exercise Set 1.3 1. (a) −∞ (b) + ∞ 2. (a) 2 (b) 0 3. (a) 0 (b) − 1 4. (a) does not exist (b) 0 5. (a) 3 + 3 · ( − 5) = − 12 (b) 0 − 4 · ( − 5) + 1 = 21 (c) 3 · ( − 5) = − 15 (d) ( − 5) 2 = 25 (e) 3 √ 5 + 3 = 2 (f ) 3 / ( − 5) = − 3 / 5 (g) 0 (h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. 6. (a) 2 · 7 − ( − 6) = 20 (b) 6 · 7 + 7 · ( − 6) = 0 (c) + ∞ (d) −∞ (e) 3 √ − 42 (f ) − 6 / 7 (g) 7 (h) − 7 / 12 7. (a) x 0 . 1 0 . 01 0 . 001 0 . 0001 0 . 00001 0 . 000001 f ( x ) 1 . 471128 1 . 560797 1 . 569796 1 . 570696 1 . 570786 1 . 570795 The limit appears to be ≈ 1 . 57079 . . . . (b) The limit is π/ 2. 8. x 10 100 1000 10000 100000 1000000 f ( x ) 1 . 258925 1 . 047129 1 . 006932 1 . 000921 1 . 000115 1 . 000014 The limit appears to be 1. 9. The limit is −∞ , by the highest degree term. 10. The limit is + ∞ , by the highest degree term. 11. The limit is + ∞ . 12. The limit is + ∞ . 13. The limit is 3 / 2, by the highest degree terms. 14. The limit is 5 / 2, by the highest degree terms. 15. The limit is 0, by the highest degree terms. 16. The limit is 0, by the highest degree terms. 17. The limit is 0, by the highest degree terms. 18. The limit is 5 / 3, by the highest degree terms. 19. The limit is −∞ , by the highest degree terms. 20. The limit is + ∞ , by the highest degree terms. 21. The limit is − 1 / 7, by the highest degree terms. 22. The limit is 4 / 7, by the highest degree terms.

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    Exercise Set 1.3 47 23. The limit is 3 √ − 5 / 8 = − 3 √ 5 / 2, by the highest degree terms. 24. The limit is 3 √ 3 / 2 , by the highest degree terms. 25. √ 5 x 2 − 2 x + 3 = √ 5 − 2 x 2 − 1 − 3 x when x < 0. The limit is − √ 5 . 26. √ 5 x 2 − 2 x + 3 = √ 5 − 2 x 2 1 + 3 x when x > 0. The limit is √ 5 . 27. 2 − y √ 7 + 6 y 2 = − 2 y + 1 √ 7 y 2 + 6 when y < 0. The limit is 1 / √ 6 . 28. 2 − y √ 7 + 6 y 2 = 2 y − 1 √ 7 y 2 + 6 when y > 0. The limit is − 1 / √ 6 . 29. √ 3 x 4 + x x 2 − 8 = √ 3 + 1 x 3 1 − 8 x 2 when x < 0. The limit is √ 3 . 30. √ 3 x 4 + x x 2 − 8 = √ 3 + 1 x 3 1 − 8 x 2 when x > 0. The limit is √ 3 . 31. lim x → + ∞ ( √ x 2 + 3 − x ) √ x 2 + 3 + x √ x 2 + 3 + x = lim x → + ∞ 3 √ x 2 + 3 + x = 0, by the highest degree terms. 32. lim x → + ∞ ( √ x 2 − 3 x − x ) √ x 2 − 3 x + x √ x 2 − 3 x + x = lim x → + ∞ − 3 x √ x 2 − 3 x + x = − 3 / 2, by the highest degree terms. 33. lim x →−∞ 1 − e x 1 + e x = 1 − 0 1 + 0 = 1. 34. Divide the numerator and denominator by e x : lim x → + ∞ 1 − e x 1 + e x = lim x → + ∞ e − x − 1 e − x + 1 = 0 − 1 0 + 1 = − 1. 35. Divide the numerator and denominator by e x : lim x → + ∞ 1 + e − 2 x 1 − e − 2 x = 1 + 0 1 − 0 = 1. 36. Divide the numerator and denominator by e − x : lim x →−∞ e 2 x + 1 e 2 x − 1 = 0 + 1 0 − 1 = − 1. 37. The limit is −∞ . 38. The limit is + ∞ . 39. x + 1 x = 1 + 1 x , so lim x → + ∞ ( x + 1) x x x = e from Figure 1.3.4. 40. ( 1 + 1 x ) − x = 1 ( 1 + 1 x ) x , so the limit is e − 1 . 41. False: lim x → + ∞ ( 1 + 1 x ) 2 x = [ lim x → + ∞ ( 1 + 1 x ) x ] 2 = e 2 . 42. False; y = 0 is a horizontal asymptote for the curve y = e x yet lim x → + ∞ e x does not exist. 43. True: for example f ( x ) = sin x/x crosses the x -axis infinitely many times at x = nπ, n = 1 , 2 , . . . .

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    48 Chapter 1 44. False: if the asymptote is y = 0, then lim x →±∞ p ( x ) /q ( x ) = 0, and clearly the degree of p ( x ) is strictly less than the degree of q ( x ). If the asymptote is y = L 6 = 0, then lim x →±∞ p ( x ) /q ( x ) = L and the degrees must be equal. 45. It appears that lim t → + ∞ n ( t ) = + ∞ , and lim t → + ∞ e ( t ) = c . 46. (a) It is the initial temperature of the potato (400 ◦ F). (b) It is the ambient temperature, i.e. the temperature of the room. 47. (a) + ∞ (b) − 5 48. (a) 0 (b) − 6 49. lim x →−∞ p ( x ) = + ∞ . When n is even, lim x → + ∞ p ( x ) = + ∞ ; when n is odd, lim x → + ∞ p ( x ) = −∞ . 50. (a) p ( x ) = q ( x ) = x . (b) p ( x ) = x , q ( x ) = x 2 . (c) p ( x ) = x 2 , q ( x ) = x . (d) p ( x ) = x + 3, q ( x ) = x . 51. (a) No. (b) Yes, tan x and sec x at x = nπ + π/ 2 and cot x and csc x at x = nπ, n = 0 , ± 1 , ± 2 , . . . . 52. If m > n the limit is zero. If m = n the limit is c m /d m . If n > m the limit is + ∞ if c n d m > 0 and −∞ if c n d m < 0. 53. (a) Every value taken by e x 2 is also taken by e t : choose t = x 2 . As x and t increase without bound, so does e t = e x 2 . Thus lim x → + ∞ e x 2 = lim t → + ∞ e t = + ∞ . (b) If f ( t ) → + ∞ (resp. f ( t ) → −∞ ) then f ( t ) can be made arbitrarily large (resp. small) by taking t large enough. But by considering the values g ( x ) where g ( x ) > t , we see that f ( g ( x )) has the limit + ∞ too (resp. limit −∞ ). If f ( t ) has the limit L as t → + ∞ the values f ( t ) can be made arbitrarily close to L by taking t large enough. But if x is large enough then g ( x ) > t and hence f ( g ( x )) is also arbitrarily close to L . (c) For lim x →−∞ the same argument holds with the substitutiion ” x decreases without bound” instead of ” x increases without bound”. For lim x → c − substitute ” x close enough to c, x < c ”, etc. 54. (a) Every value taken by e − x 2 is also taken by e t : choose t = − x 2 . As x increases without bound and t decreases without bound, the quantity e t = e − x 2 tends to 0. Thus lim x → + ∞ e − x 2 = lim t →−∞ e t = 0. (b) If f ( t ) → + ∞ (resp. f ( t ) → −∞ ) then f ( t ) can be made arbitrarily large (resp. small) by taking t small enough. But by considering the values g ( x ) where g ( x ) < t , we see that f ( g ( x )) has the limit + ∞ too (resp. limit −∞ ). If f ( t ) has the limit L as t → −∞ the values f ( t ) can be made arbitrarily close to L by taking t small enough. But if x is large enough then g ( x ) < t and hence f ( g ( x )) is also arbitrarily close to L . (c) For lim x →−∞ the same argument holds with the substitutiion ” x decreases without bound” instead of ” x increases without bound”. For lim x → c − substitute ” x close enough to c, x < c ”, etc. 55. t = 1 /x, lim t → + ∞ f ( t ) = + ∞ . 56. t = 1 /x, lim t →−∞ f ( t ) = 0. 57. t = csc x, lim t → + ∞ f ( t ) = + ∞ . 58. t = csc x, lim t →−∞ f ( t ) = 0. 59. Let t = ln x . Then t also tends to + ∞ , and ln 2 x ln 3 x = t + ln 2 t + ln 3 , so the limit is 1. 60. With t = x − 1 , [ln( x 2 − 1) − ln( x + 1)] = ln( x + 1) + ln( x − 1) − ln( x + 1) = ln t , so the limit is + ∞ .

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    Exercise Set 1.3 49 61. Set t = − x , then get lim t →−∞ ( 1 + 1 t ) t = e by Figure 1.3.4. 62. With t = x/ 2, lim x → + ∞ ( 1 + 2 x ) x = ( lim t → + ∞ [1 + 1 /t ] t ) 2 = e 2 63. From the hint, lim x → + ∞ b x = lim x → + ∞ e (ln b ) x =        0 if b < 1 , 1 if b = 1 , + ∞ if b > 1 . 64. It suffices by Theorem 1.1.3 to show that the left and right limits at zero are equal to e . (a) lim x → + ∞ (1 + x ) 1 /x = lim t → 0 + (1 + 1 /t ) t = e . (b) lim x →−∞ (1 + x ) 1 /x = lim t → 0 − (1 + 1 /t ) t = e . 65. (a) 4 8 12 16 20 40 80 120 160 200 t v (b) lim t →∞ v = 190 ( 1 − lim t →∞ e − 0 . 168 t ) = 190, so the asymptote is v = c = 190 ft/sec. (c) Due to air resistance (and other factors) this is the maximum speed that a sky diver can attain. 66. (a) p (1990) = 525 / (1 + 1 . 1) = 250 (million). (b) 1920 2000 2080 250 500 t P (c) lim t →∞ p ( t ) = 525 1 + 1 . 1 lim t →∞ e − 0 . 02225( t − 1990) = 525 (million). (d) The population becomes stable at this number. 67. (a) n 2 3 4 5 6 7 1 + 10 − n 1 . 01 1 . 001 1 . 0001 1 . 00001 1 . 000001 1 . 0000001 1 + 10 n 101 1001 10001 100001 1000001 10000001 (1 + 10 − n ) 1+10 n 2 . 7319 2 . 7196 2 . 7184 2 . 7183 2 . 71828 2 . 718282 The limit appears to be e . (b) This is evident from the lower left term in the chart in part (a). (c) The exponents are being multiplied by a , so the result is e a .

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    50 Chapter 1 68. (a) f ( − x ) = ( 1 − 1 x ) − x = ( x − 1 x ) − x = ( x x − 1 ) x , f ( x − 1) = ( x x − 1 ) x − 1 = ( x − 1 x ) f ( − x ). (b) lim x →−∞ ( 1 + 1 x ) x = lim x → + ∞ f ( − x ) = [ lim x → + ∞ x x − 1 ] lim x → + ∞ f ( x − 1) = lim x → + ∞ f ( x − 1) = e . 69. After a long division, f ( x ) = x + 2 + 2 x − 2 , so lim x →±∞ ( f ( x ) − ( x + 2)) = 0 and f ( x ) is asymptotic to y = x + 2. The only vertical asymptote is at x = 2. –12 –6 3 9 15 –15 –9 –3 3 9 15 x x = 2 y y = x + 2 70. After a simplification, f ( x ) = x 2 − 1 + 3 x , so lim x →±∞ ( f ( x ) − ( x 2 − 1)) = 0 and f ( x ) is asymptotic to y = x 2 − 1. The only vertical asymptote is at x = 0. –4 –2 2 4 –2 1 3 5 x y y = x 2 – 1 71. After a long division, f ( x ) = − x 2 +1+ 2 x − 3 , so lim x →±∞ ( f ( x ) − ( − x 2 +1)) = 0 and f ( x ) is asymptotic to y = − x 2 +1. The only vertical asymptote is at x = 3. –4 –2 2 4 –12 –6 6 12 x x = 3 y y = – x 2 + 1 72. After a long division, f ( x ) = x 3 + 3 2( x − 1) − 3 2( x + 1) , so lim x →±∞ ( f ( x ) − x 3 ) = 0 and f ( x ) is asymptotic to y = x 3 . The vertical asymptotes are at x = ± 1.

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    Exercise Set 1.4 51 –2 2 –15 5 15 x y y = x 3 x = 1 x = –1 73. lim x →±∞ ( f ( x ) − sin x ) = 0 so f ( x ) is asymptotic to y = sin x . The only vertical asymptote is at x = 1. –4 2 8 –4 3 5 x y y = sin x x = 1 Exercise Set 1.4 1. (a) | f ( x ) − f (0) | = | x + 2 − 2 | = | x | < 0 . 1 if and only if | x | < 0 . 1. (b) | f ( x ) − f (3) | = | (4 x − 5) − 7 | = 4 | x − 3 | < 0 . 1 if and only if | x − 3 | < (0 . 1) / 4 = 0 . 025. (c) | f ( x ) − f (4) | = | x 2 − 16 | <  if | x − 4 | < δ . We get f ( x ) = 16 +  = 16 . 001 at x = 4 . 000124998, which corresponds to δ = 0 . 000124998; and f ( x ) = 16 −  = 15 . 999 at x = 3 . 999874998, for which δ = 0 . 000125002. Use the smaller δ : thus | f ( x ) − 16 | <  provided | x − 4 | < 0 . 000125 (to six decimals). 2. (a) | f ( x ) − f (0) | = | 2 x + 3 − 3 | = 2 | x | < 0 . 1 if and only if | x | < 0 . 05. (b) | f ( x ) − f (0) | = | 2 x + 3 − 3 | = 2 | x | < 0 . 01 if and only if | x | < 0 . 005. (c) | f ( x ) − f (0) | = | 2 x + 3 − 3 | = 2 | x | < 0 . 0012 if and only if | x | < 0 . 0006. 3. (a) x 0 = (1 . 95) 2 = 3 . 8025 , x 1 = (2 . 05) 2 = 4 . 2025. (b) δ = min ( | 4 − 3 . 8025 | , | 4 − 4 . 2025 | ) = 0 . 1975. 4. (a) x 0 = 1 / (1 . 1) = 0 . 909090 . . . , x 1 = 1 / (0 . 9) = 1 . 111111 . . . (b) δ = min( | 1 − 0 . 909090 | , | 1 − 1 . 111111 | ) = 0 . 0909090 . . . 5. | ( x 3 − 4 x +5) − 2 | < 0 . 05 is equivalent to − 0 . 05 < ( x 3 − 4 x +5) − 2 < 0 . 05, which means 1 . 95 < x 3 − 4 x +5 < 2 . 05. Now x 3 − 4 x +5 = 1 . 95 at x = 1 . 0616, and x 3 − 4 x +5 = 2 . 05 at x = 0 . 9558. So δ = min (1 . 0616 − 1 , 1 − 0 . 9558) = 0 . 0442.

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    52 Chapter 1 2.2 1.9 0.9 1.1 6. √ 5 x + 1 = 3 . 5 at x = 2 . 25, √ 5 x + 1 = 4 . 5 at x = 3 . 85, so δ = min(3 − 2 . 25 , 3 . 85 − 3) = 0 . 75. 5 0 2 4 7. With the TRACE feature of a calculator we discover that (to five decimal places) (0 . 87000 , 1 . 80274) and (1 . 13000 , 2 . 19301) belong to the graph. Set x 0 = 0 . 87 and x 1 = 1 . 13. Since the graph of f ( x ) rises from left to right, we see that if x 0 < x < x 1 then 1 . 80274 < f ( x ) < 2 . 19301, and therefore 1 . 8 < f ( x ) < 2 . 2. So we can take δ = 0 . 13. 8. From a calculator plot we conjecture that lim x → 0 f ( x ) = 2. Using the TRACE feature we see that the points ( ± 0 . 2 , 1 . 94709) belong to the graph. Thus if − 0 . 2 < x < 0 . 2, then 1 . 95 < f ( x ) ≤ 2 and hence | f ( x ) − L | < 0 . 05 < 0 . 1 =  . 9. | 2 x − 8 | = 2 | x − 4 | < 0 . 1 when | x − 4 | < 0 . 1 / 2 = 0 . 05 = δ . 10. | (5 x − 2) − 13 | = 5 | x − 3 | < 0 . 01 when | x − 3 | < 0 . 01 / 5 = 0 . 002 = δ . 11. If x 6 = 3, then ∣ ∣ ∣ ∣ x 2 − 9 x − 3 − 6 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ x 2 − 9 − 6 x + 18 x − 3 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ x 2 − 6 x + 9 x − 3 ∣ ∣ ∣ ∣ = | x − 3 | < 0 . 05 when | x − 3 | < 0 . 05 = δ . 12. If x 6 = − 1 / 2, then ∣ ∣ ∣ ∣ 4 x 2 − 1 2 x + 1 − ( − 2) ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 4 x 2 − 1 + 4 x + 2 2 x + 1 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 4 x 2 + 4 x + 1 2 x + 1 ∣ ∣ ∣ ∣ = | 2 x + 1 | = 2 | x − ( − 1 / 2) | < 0 . 05 when | x − ( − 1 / 2) | < 0 . 025 = δ . 13. Assume δ ≤ 1. Then − 1 < x − 2 < 1 means 1 < x < 3 and then | x 3 − 8 | = | ( x − 2)( x 2 + 2 x + 4) | < 19 | x − 2 | , so we can choose δ = 0 . 001 / 19. 14. Assume δ ≤ 1. Then − 1 < x − 4 < 1 means 3 < x < 5 and then | √ x − 2 | = ∣ ∣ ∣ ∣ x − 4 √ x + 2 ∣ ∣ ∣ ∣ < | x − 4 | √ 3 + 2 , so we can choose δ = 0 . 001 · ( √ 3 + 2). 15. Assume δ ≤ 1. Then − 1 < x − 5 < 1 means 4 < x < 6 and then ∣ ∣ ∣ ∣ 1 x − 1 5 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ x − 5 5 x ∣ ∣ ∣ ∣ < | x − 5 | 20 , so we can choose δ = 0 . 05 · 20 = 1. 16. || x | − 0 | = | x | < 0 . 05 when | x − 0 | < 0 . 05 = δ . 17. Let  > 0 be given. Then | f ( x ) − 3 | = | 3 − 3 | = 0 <  regardless of x , and hence any δ > 0 will work. 18. Let  > 0 be given. Then | ( x + 2) − 6 | = | x − 4 | <  provided δ =  (although any smaller δ would work).

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    Exercise Set 1.4 53 19. | 3 x − 15 | = 3 | x − 5 | <  if | x − 5 | < / 3, δ = / 3. 20. | 7 x + 5 + 2 | = 7 | x + 1 | <  if | x + 1 | < / 7, δ = / 7. 21. ∣ ∣ ∣ ∣ 2 x 2 + x x − 1 ∣ ∣ ∣ ∣ = | 2 x | <  if | x | < / 2, δ = / 2. 22. ∣ ∣ ∣ ∣ x 2 − 9 x + 3 − ( − 6) ∣ ∣ ∣ ∣ = | x + 3 | <  if | x + 3 | <  , δ =  . 23. | f ( x ) − 3 | = | x + 2 − 3 | = | x − 1 | <  if 0 < | x − 1 | <  , δ =  . 24. | 9 − 2 x − 5 | = 2 | x − 2 | <  if 0 < | x − 2 | < / 2, δ = / 2. 25. If  > 0 is given, then take δ =  ; if | x − 0 | = | x | < δ , then | x − 0 | = | x | <  . 26. If x < 2 then | f ( x ) − 5 | = | 9 − 2 x − 5 | = 2 | x − 2 | <  if | x − 2 | < / 2 , δ 1 = / 2. If x > 2 then | f ( x ) − 5 | = | 3 x − 1 − 5 | = 3 | x − 2 | <  if | x − 2 | < / 3 , δ 2 = / 3 Now let δ = min( δ 1 , δ 2 ) then for any x with | x − 2 | < δ, | f ( x ) − 5 | <  . 27. For the first part, let  > 0. Then there exists δ > 0 such that if a < x < a + δ then | f ( x ) − L | <  . For the left limit replace a < x < a + δ with a − δ < x < a . 28. (a) Given  > 0 there exists δ > 0 such that if 0 < | x − a | < δ then || f ( x ) − L | − 0 | <  , or | f ( x ) − L | <  . (b) From part (a) it follows that | f ( x ) − L | <  is the defining condition for each of the two limits, so the two limit statements are equivalent. 29. (a) | (3 x 2 + 2 x − 20 − 300 | = | 3 x 2 + 2 x − 320 | = | (3 x + 32)( x − 10) | = | 3 x + 32 | · | x − 10 | . (b) If | x − 10 | < 1 then | 3 x + 32 | < 65, since clearly x < 11. (c) δ = min(1 , / 65); | 3 x + 32 | · | x − 10 | < 65 · | x − 10 | < 65 · / 65 =  . 30. (a) ∣ ∣ ∣ ∣ 28 3 x + 1 − 4 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 28 − 12 x − 4 3 x + 1 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ − 12 x + 24 3 x + 1 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 12 3 x + 1 ∣ ∣ ∣ ∣ · | x − 2 | . (b) If | x − 2 | < 4 then − 2 < x < 6, so x can be very close to − 1 / 3, hence ∣ ∣ ∣ ∣ 12 3 x + 1 ∣ ∣ ∣ ∣ is not bounded. (c) If | x − 2 | < 1 then 1 < x < 3 and 3 x + 1 > 4, so ∣ ∣ ∣ ∣ 12 3 x + 1 ∣ ∣ ∣ ∣ < 12 4 = 3. (d) δ = min(1 , / 3); ∣ ∣ ∣ ∣ 12 3 x + 1 ∣ ∣ ∣ ∣ · | x − 2 | < 3 · | x − 2 | < 3 · / 3 =  . 31. If δ < 1 then | 2 x 2 − 2 | = 2 | x − 1 || x + 1 | < 6 | x − 1 | <  if | x − 1 | < / 6, so δ = min(1 , / 6). 32. If δ < 1 then | x 2 + x − 12 | = | x + 4 | · | x − 3 | < 5 | x − 3 | <  if | x − 3 | < / 5, so δ = min(1 , / 5). 33. If δ < 1 / 2 and | x − ( − 2) | < δ then − 5 / 2 < x < − 3 / 2, x + 1 < − 1 / 2, | x + 1 | > 1 / 2; then ∣ ∣ ∣ ∣ 1 x + 1 − ( − 1) ∣ ∣ ∣ ∣ = | x + 2 | | x + 1 | < 2 | x + 2 | <  if | x + 2 | < / 2, so δ = min(1 / 2 , / 2). 34. If δ < 1 / 4 and | x − (1 / 2) | < δ then ∣ ∣ ∣ ∣ 2 x + 3 x − 8 ∣ ∣ ∣ ∣ = | 6 x − 3 | | x | < 6 | x − (1 / 2) | 1 / 4 = 24 | x − (1 / 2) | <  if | x − (1 / 2) | < / 24 , so δ = min(1 / 4 , / 24).

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