Instructor's Solutions Manual for Introduction to Analysis, by William R. Wade

By William R. Wade

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By William R. Wade

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X−a x−a 29 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2, y(x) → nan−1 as x → a. Since z(x) contains m terms, it is easy to see that z(x) → maq(m−1) as x → a. Therefore, xq is differentiable at a and the value of its derivative there is nan−1 · (maq(m−1) )−1 = qan−1−qm+q = qaq−1 . 8. Clearly, f (x) exists when x = 0. By definition, h/(1 + e1/h ) − 0 1 = lim = 0. h→0+ h→0+ 1 + e1/h h f (0) = lim Since this limit is 1 as h → 0−, f is not differentiable at x = 0. 9. a) By assumptions ii) and vi), 0 ≤ | sin x| ≤ |x| for x ∈ [−π/2, π/2].

C) True. Since f is continuous, it is locally integrable on (a, b). If f (x) < 1, then f (x) < 1 ≤ 1 + f (x). If f (x) ≥ 1, then f (x) ≤ f (x) < 1 + f (x). Thus f (x) < 1 + f (x) for all x ∈√(a, b). The function 1 + f is absolutely integrable on (a, b) by hypothesis and the fact that b − a < ∞. Thus f is absolutely integrable on (a, b) by the Comparison Theorem. d) True. 42 and the Comparison Theorem that any finite linear combination of f, g, and |f − g| is absolutely integrable on (a, b). 18, f ∨ g and f ∧ g are absolutely integrable on (a, b).

By L’Hˆopital’s Rule, Bα → 1 as α → 0+ and Bα → 0 as α → ∞. 7. a) f (x) = 2e−1/x /x3 is evidently continuous for x = 0. Also, by L’Hˆopital’s Rule, limx→0 f (x) = 2 2 2 2 limx→0 (2/x3 )/e1/x = limx→0 (6/x4 )/(2e1/x /x3 ) = limx→0 (3/x)/e1/x = limx→0 (3/x2 )/(2e1/x /x3 ) = 0. On the 2 2 other hand, f (0) := limh→0 (e−1/h − 0)/h = limh→0 (1/h)/e1/h = 0. Thus f exists and is continuous on R. 2 b) We first prove that the functions g(x) = e−1/x /xk satisfy g(x) → 0 as x → 0 for all integers k ≥ 0. Indeed, this surely holds for k = 0.

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