ans0918 - Homework answers week of Sept 18#7 p 41 Since f(x...

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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 0 < | 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 0 | x | = 0, the sandwich theorem im- plies that lim x 0 | f ( x ) | = 0. Hence lim x 0 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 ) = 0 . Clearly f ( 0 ) is undefined, because no reasonable value can be as- signed to sin ( 1 / 0 ) . Therefore, f ( g ( 1 / n π )) = f ( f ( 1 / n π )) = f ( 0 ) , which is undefined. (c) The definition (p. 34) of the limit of a function as x p requires, among other things, that f be defined on some interval around p (though not necessarily at p itself). This is so that it is reasonable to talk about | f ( x ) - f ( p ) | for all values of x
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