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FA07Ma1aSol8

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

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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 0 (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 0 (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 ) 0 = x x (log x + 1) - 1 and (1 - x + log x ) 0 = - 1 + 1 /x. Note that both derivatives tend to 0 when x 1+. We shall compute the deriva- tives of them again. ( x x ) 00 = x x ((log x + 1) 2 + 1 /x ) and (1 - x + log x ) 00 = - 1 /x 2 . Then by L’Hˆ opital’s rule, lim x 1 + x x - x 1 - x + log x = lim x 1 + x x ((log x + 1) 2 + 1 x ) - 1 x 2 = - 2 .

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

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