# A given any m 0 there corresponds n 0 such that if x

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Unformatted text preview: 2| > 200, −x − 2 > 200, x < −202, N = −202 x+2 38. 1 < 0.01 if |x| > 10, −x > 10, x < −10, N = −10 x2 39. 11 4x − 1 −2 = < 0.1 if |2x + 5| > 110, −2x − 5 > 110, 2x < −115, x < −57.5, N = −57.5 2x + 5 2x + 5 40. 1 x −1 = < 0.001 if |x + 1| > 1000, −x − 1 > 1000, x < −1001, N = −1001 x+1 x+1 41. 1 < x2 43. 1 < x+2 if |x + 2| > 1 44. 1 < x+2 if |x + 2| > 1 45. x 1 −1 = < x+1 x+1 if |x + 1| > 1 46. x 1 −1 = < x+1 x+1 if |x + 1| > 1 47. 4x − 1 11 11 11 11 11 5 5 11 −2 = < if |2x +5| > , −2x − 5 > , 2x < − − 5, x < − − , N = − − 2x + 5 2x + 5 2 2 22 48. 4x − 1 11 −2 = < 2x + 5 2x + 5 49. (a) (c) 1 1 if |x| > √ , N = √ , −x − 2 < , x+2> 1 < x 42. 1 1 51. 1 , −x > 1 1 1 ,x<− ,N =− 1 1 , x > −2 − , N = −2 − ,x> 1 − 2, N = ,x> 1 11 −2 − 1, N = , −x − 1 > if |2x + 5| > 1 , 2x > 1 1 > 100 if |x| < 2 x 10 −1 1 < −1000 if |x − 3| < √ (x − 3)2 10 10 1 1 −1 1 1 , x < −1 − , N = −1 − 11 − 5, x > (b) (d) (a) 1 1 > 10 if and only if |x − 1| < √ (x − 1)2 10 (c) 50. if |x| > 11 5 11 5 − ,N= − 2 2 2 2 1 1 > 1000 if |x − 1| < |x − 1| 1000 1 1 1 , |x| < − 4 < −10000 if x4 < x 10000 10 1 1 √ > 100000 if and only if |x − 1| < 2 (x − 1) 100 10 if M > 0 then (b) 1 > 1000 if and only if (x − 1)2 1 |x − 1| < √ 10 10 1 1 1 1 , 0 < |x − 3| < √ , δ = √ > M , 0 < (x − 3)2 < 2 (x − 3) M M M 53 Chapter 2 52. if M < 0 then 1 −1 1 1 ,δ=√ < M , 0 < (x − 3)2 < − , 0 < |x − 3| < √ (x − 3)2 M −M −M 53. if M > 0 then 1 1 1 > M , 0 < |x| < ,δ= |x| M M 54. if M > 0 then 1 1 1 > M , 0 < |x − 1| < ,δ= |x − 1| M M 55. if M < 0 then − 56. if M > 0 then 57. if x > 2 then |x + 1 − 3| = |x − 2| = x − 2 < 58. if x < 1 then |3x + 2 − 5| = |3x − 3| = 3|x − 1| = 3(1 − x) < 59. if x > 4 then 60. if x < 0 then 61. if x > 2 then |f (x) − 2| = |x − 2| = x − 2 < 62. if x < 2 then |f (x) − 6| = |3x − 6| = 3|x − 2| = 3(2 − x) < 63. 64. 1 1 1 1 < M , 0 < x4 < − , |x| < ,δ= 4 1/4 x M (−M ) (−M )1/4 1 1 1 1 , x < 1/4 , δ = 1/4 > M , 0 < x4 < 4 x M M M √ √ x−4< −x < if x − 4 < if −x < 2 ,− 2 if 2 < x < 2 + , δ = , 4<x<4+ 2 < x < 0, δ = 2 ,δ= if 1 − x < 1 3 , 1− 1 3 < x < 1, δ = 1 3 2 2 if 2 < x < 2 + , δ = if 2 − x < 1 3 , 2− 1 3 < x < 2, δ = 1 3 1 1 1 1 < M, x − 1 < − , 1 < x < 1 − ,δ=− 1−x M M M 1 1 1 1 (b) if M > 0 and x < 1 then > M, 1 − x < , 1− < x < 1, δ = 1−x M M M (a) if M < 0 and x > 1 then 1 1 1 1 > M, x < ,0<x< ,δ= x M M M 11 1 1 < x < 0, δ = − (b) if M < 0 and x < 0 then < M , −x < − , x MM M 66. if M > 0 and x > 0 then (a) Given any M > 0 there corresponds N > 0 such that if x > N then f (x) > M , x + 1 > M , x > M − 1, N = M − 1. (b) 65. (a) Given any M < 0 there corresponds N < 0 such that if x < N then f (x) < M , x + 1 < M , x < M − 1, N = M − 1. (a) Given any M > 0 there corresponds N > 0 such that if x > N then f (x) > M , x2 − 3 > M , √ √ x > M + 3, N = M + 3. (b) Given any M < 0 there corresponds N < 0 such that if x < N then f (x) < M , x3 + 5 < M , x < (M − 5)1/3 , N = (M − 5)1/3 . 67. if δ ≤ 2 then |x − 3| < 2, −2 < x − 3 < 2, 1 < x < 5, and |x2 − 9| = |x + 3||x − 3| < 8|x − 3| < |x − 3| < 1 , δ = min 2, 1 8 8 68. (a) (b) if We don’t care about the value of f at x = a, because the limit is only concerned with values of x near a. The condition that f be deﬁned for all x (except possibly x = a) is necessary, because if some points were excluded then the limit may not exist; for example, let f (x) = x if 1/x is not an integer and f (1/n) = 6. Then lim f (x) does not exist but it would if the points 1/n were excluded. √ when x < 0 then x is not deﬁned x→0 (c) yes; if δ ≤ 0.01 then x > 0, so √ x is deﬁned Exercise Set 2.4 69. 54 |(x3 − 4x +5) − 2| < 0.05, −0.05 < (x3 − 4x + 5) − 2 < 0.05, 1.95 < x3 − 4x + 5 < 2.05; x3 − 4x + 5 = 1.95 at x = 1.0616, x3 − 4x + 5 = 2.05 at x = 0.9558; δ = min (1.0616 − 1, 1 − 0.9558) = 0.0442 2.2 0.9 1.9 70. √ 1.1 5x + 1 = 3.5 at x = 2.25, √ 5x + 1 = 4.5 at x = 3.85, so δ = min(3 − 2.25, 3.85 − 3) = 0.75 5 2 4 0 EXERCISE SET 2.4 1. (a) no, x = 2 (b) no, x = 2 (e) yes (f ) yes (a) no, x = 2 (b) no, x = 2 (e) no, x = 2 (f ) yes (a) no, x = 1, 3 (b) yes (e) no, x = 3 (f ) yes (a) no, x = 3 (b) yes (e) no, x = 3 (f ) yes 5. (a) 3 (b) 3 6. −2/5 7. (a) 2. 3. 4. (c) no, x = 2 (c) (d) yes...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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