liggett_quizsolutions3,4

liggett_quizsolutions3,4 - Mathematics 31B/4 Fall 2004 Quiz...

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Quiz for Tuesday, November 2 1. Evaluate (a) lim x 0 e x - 1 - x x 2 Solution: This is an indeterminate form, so we may apply L’Hospital’s rule, to get lim x 0 e x - 1 - x x 2 = lim x 0 e x - 1 2 x if the latter limit exists. This again is indeterminate, so we may apply L’Hospital’s rule again to get lim x 0 e x - 1 2 x = lim x 0 e x 2 = 1 2 . (b) lim x 0 (1 - 2 x ) 1 /x Solution: Taking logs, we see that we must evaluate lim x 0 ln(1 - 2 x ) x . This is an indeterminate form, so lim x 0 ln(1 - 2 x ) x = lim x 0 - 2 1 - 2 x = - 2 . Therefore lim x 0 (1 - 2 x ) 1 /x = e - 2 . 2. Evaluate Z 2 1 x 4 (ln x ) 2 dx Solution: Use integration by parts twice. Z 2 1 x 4 (ln x ) 2 dx = x 5 5 (ln x ) 2 2 1 - 2 5 Z 2 1 x 4 ln xdx = 32 5 (ln2) 2 - 2 25 x 5 ln x 2 1 + 2 25 Z 2 1 x 4 dx = 32 5 (ln2) 2 - 64 25 ln2 + 62 125 . Quiz for Thursday, November 4
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liggett_quizsolutions3,4 - Mathematics 31B/4 Fall 2004 Quiz...

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