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ans0918 - Homework answers, week of Sept. 18 #7, p. 41...

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Unformatted text preview: Homework answers, week of Sept. 18 #7, p. 41 Since f ( x ) and h ( x ) are continuous at a , then lim x → a f ( x ) = f ( a ) and lim x → a h ( x ) = h ( a ) . By hypothesis, f ( a ) = h ( a ) = L . Since f ( x ) ≤ g ( x ) ≤ h ( x ) for all | x- a | < k , L = f ( a ) ≤ g ( a ) ≤ h ( a ) = L , so g ( a ) = L . Furthermore, the sandwich theorem implies that lim x → a g ( x ) = L . Since the limit exists, then given ε > 0, there exists 0 < δ < k such that | g ( x )- g ( a ) | < ε whenever < | x- a | < δ . Hence g is continuous at a . #11, p. 42 Let f ( x ) = x sin ( 1 / x ) . (a) Since 0 ≤ | f ( x ) | ≤ | x | and lim x → | x | = 0, the sandwich theorem im- plies that lim x → | f ( x ) | = 0. Hence lim x → f ( x ) = 0. (b) Let n be any nonzero integer. Then f 1 n π = 1 n π sin 1 1 / n π = 1 n π sin ( n π ) = . Clearly f ( ) is undefined, because no reasonable value can be as- signed to sin ( 1 / ) . Therefore, f ( g ( 1 / n π )) = f ( f ( 1 / n π )) = f (...
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This note was uploaded on 04/02/2009 for the course MAT 371 taught by Professor Thieme during the Spring '07 term at ASU.

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ans0918 - Homework answers, week of Sept. 18 #7, p. 41...

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