Unformatted text preview: ) f ( c )  ≤ ε ( vu ). [Hint: Add and substract f ( c )cf ( c ) to the left side.] Fix ε > 0. Since f is diﬀerentiable at c , we have that for any x ∈ I with  xc  < δ ± ± ± ± f ( x )f ( c ) xcf ( c ) ± ± ± ± < ε/ 2 . That is  f ( x )f ( c )f ( c )( xc ) < ε 2 ( xc ) . We apply this with x = u so that cδ < u ≤ c and x = v with c ≤ v < c + δ . Then,  f ( v )f ( u )( vu ) f ( c )  ≤  f ( v )f ( c )( vc ) f ( c )  +  f ( u )f ( c )( uc ) f ( c )  ≤ ε 2  vc  + ε 2  uc  ≤ ε  vu  . 1...
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 MetCalfe
 Calculus

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