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Unformatted text preview: Math 125A, Fall 2009 Solutions for Homework 4 (Prepared by Matt Low) 1. Regardless of the argument of the cosine function, we have  cos u  ≤ 1, and so: lim x → x cos ( 3 x 2 + e sin x ln  x  ) ≤ lim x →  x  = 0 and consequently: lim x → x cos ( 3 x 2 + e sin x ln  x  ) = 0 2. (a) False. For example, the function f ( x ) = sin( πx ) is continuous and satisfies: lim n →∞ a n = lim n →∞ sin( nπ ) = lim n →∞ 0 = 0 But lim x →∞ f ( x ) does not even exist. (b) False. For example, for the functions f ( x ) = x and g ( x ) = x , we have f ( x ) → ∞ and g ( x ) → ∞ as x → ∞ , and: lim x →∞ [ f ( x ) + g ( x )] = lim x →∞ 0 = 0 (c) False. Likewise, we can take f ( x ) = x and g ( x ) = 1 /x . Of course, we have f ( x ) → ∞ and g ( x ) → 0 as x → ∞ , and: lim x →∞ f ( x ) g ( x ) = lim x →∞ 1 = 1 3. (a) Let ε > 0 be given. In accordance with statement (i), there is a δ > 0 so that if  h  < δ , then  f ( x + h ) f...
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This note was uploaded on 03/18/2012 for the course MATH 67 taught by Professor Schilling during the Fall '07 term at UC Davis.
 Fall '07
 Schilling

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