FA07Ma1aSol8

FA07Ma1aSol8 - MA 1A FALL 2007 HOMEWORK 8 SOLUTIONS Problem...

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Unformatted text preview: MA 1A FALL 2007 HOMEWORK 8 SOLUTIONS Problem 1. Apostol, page 295, problems 1, 4, 8, 11. Evaluate the limits: 1. lim x 2 3 x 2 + 2 x- 16 x 2- x- 2 . Solution. Since 3 x 2 + 2 x- 16 = ( x- 2)(3 x + 8) and x 2- x- 2 = ( x- 2)( x + 1), we have lim x 2 3 x 2 + 2 x- 16 x 2- x- 2 = lim x 2 3 x + 8 x + 1 = lim x 2 (3 x + 8) lim x 2 ( x + 1) = 14 3 . 4. lim x (2- x ) e x- x- 2 x 3 . Solution. Since e x = X n =0 x n n ! = 1 + x + x 2 2 + x 3 6 + o ( x 3 ) , we have (2- x ) e x- x- 2 = 2 + 2 x + x 2 + x 3 3- x- x 2- x 3 2- x- x + o ( x 3 ) =- x 3 6 + o ( x 3 ) . Therefore, lim x (2- x ) e x- x- 2 x 3 =- 1 6 . 8. lim x 1 + x x- x 1- x + log x . Solution. First note that when x 1 + , both the numerator and the denominator tend to 0. We shall compute the derivatives as follows. ( x x- x ) = x x (log x + 1)- 1 and (1- x + log x ) =- 1 + 1 /x. Note that both derivatives tend to 0 when x 1+. We shall compute the deriva- tives of them again....
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This note was uploaded on 02/22/2010 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.

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FA07Ma1aSol8 - MA 1A FALL 2007 HOMEWORK 8 SOLUTIONS Problem...

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