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Unformatted text preview: Math 125A, Fall 2009 Solutions for Homework 3 (Prepared by Matt Low) 1. Let > 0 be given. By the definition of uniform continuity, there exists a number > 0 with the property that if  a b  < then  f ( a ) f ( b )  < . Because  x n y n  0, we may choose N so that  x n y n  < for n N . Thus, if n N , then  x n y n  < , and so  f ( x n ) f ( y n )  < . 2. (a) Note that f consists of sums and compositions of continuous functions. Per Theorems 17.4 and 17.5, this means that f itself is continuous. It follows from Theorem 19.2 that f is uniformly continuous. (b) Define two sequences ( x n ) , ( y n ) by x n = 1 n and y n = 1 n 2 . It is clear that  x n y n  0, but: lim n  f ( x n ) f ( y n )  = lim n ln n = Because  f ( x n ) f ( y n )  6 0, this contradicts the property derived in Problem 1, and so we conclude that f is not uniformly continuous. Or: Since lim x 0+ ln( x ) = , which is not a finite number, one cannot extend ln( x ) to be a continuous function on [0 , 1]. By the continuous extension theorem (theorem 19.5), ln1]....
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 Fall '07
 Schilling

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