asst6-f11_solutions

asst6-f11_solutions - MATH 137 Assignment #6 Solutions 1....

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Unformatted text preview: MATH 137 Assignment #6 Solutions 1. a) lim x e sin x- 1 x- To express this as a derivative, notice that since sin = 0, we can express the limit as lim x e sin x- e sin x- = lim x f ( x )- f ( ) x- = f ( ) where f ( x ) = e sin x The derivative of f ( x ) is calculated using the chain rule: f ( x ) = e sin x cos x = f ( ) = e sin cos = e (- 1) =- 1 Therefore, lim x e sin x- 1 x- = f ( ) =- 1 b) lim x 2- 5 32- x x If we let f ( x ) = 5 32- x , we notice that f (0) = 2 and the limit is of the form lim x 2- 5 32- x x = lim x f (0)- f ( x ) x- =- lim x f ( x )- f (0) x- =- f (0) Differentiating f ( x ) = 5 32- x : f ( x ) = 1 5 (32- x )- 4 / 5 (- 1) =- 1 5 (32- x )- 4 / 5 = f (0) =- 1 5 (32)- 4 / 5 =- 1 5 1 32 4 / 5 =- 1 80 lim x 2- 5 32- x x =- f (0) = 1 80 2. a) If H ( x ) = g ( f ( x )), then H ( x ) = g ( f ( x )) f ( x ). Thus, T ( x ) = H ( a ) + H ( a )( x- a ) = g ( f ( a )) + g ( f ( a )) f ( a )( x- a ) b) Since the tangent line to f at a is given by L ( x ) = f ( a ) + f ( a )( x- a ), and the tangent line to g at b is given by M ( y ) = g ( b ) + g ( b )( y- b ), then M ( L ( x )) = g ( b ) + g ( b )( L ( x )- b ) = g ( b ) + g ( b )( f ( a ) + f ( a )( x- a )- b ) 1 c) Substituting...
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This note was uploaded on 12/31/2011 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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asst6-f11_solutions - MATH 137 Assignment #6 Solutions 1....

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