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Unformatted text preview: ANALYSIS EXERCISE 10SOLUTIONS 1. (a) By the definition ( > 0)( > 0)( x D ( f )) (0 <  x a  < ) (  f ( x ) f ( a )  < ) . Note that  f ( x )    f ( a )   f ( x ) f ( a )  . Hence ( > 0)( > 0)( x D ( f )) (0 <  x a  < ) (  f ( x )    f ( a )  < ) . (b) Note that max { f,g } = 1 2 ( f + g ) + 1 2  f g  , min { f,g } = 1 2 ( f + g ) 1 2  f g  . The statement then follows from 1(a) and 2(a). 2. Let us prove, for instance,(b). By the definition of continuiuty we know that lim x a f ( x ) = f ( a ) , lim x a g ( x ) = g ( a ). Hence by the proper ties of the limit lim x a [ f ( x ) g ( x )] = lim x a f ( x ) lim x a g ( x ) = f ( a ) g ( a ) , which proves (b). 3. Note that  f (0)  0, hence f (0) = 0. We have to prove that lim x f ( x ) = 0. This follows from the inequality  f ( x )  M  x  by the sandwich rule....
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Winter '10 term at Bristol Community College.
 Winter '10
 Spyros

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