# sole10 - ANALYSIS EXERCISE 10SOLUTIONS 1. (a) By the...

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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.

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sole10 - ANALYSIS EXERCISE 10SOLUTIONS 1. (a) By the...

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