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midterm-2-solns

midterm-2-solns - Math 135A Winter 2008 Midterm#2(Solutions...

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Math 135A, Winter 2008 Midterm #2 (Solutions) (1) Evaluate lim x 0 cosh x - cos x sin x 2 . Solution: cosh x - cos x sin x 2 = (1 + x 2 / 2! + O ( x 3 )) - (1 - x 2 / 2! + O ( x 3 ) x 2 + O ( x 3 ) = x 2 + O ( x 3 ) x 2 + O ( x 3 ) 1 as x 0. (2) Is the series X k =1 ( - 1) k ln k k 3 + ln k + 1 absolutely convergent, conditionally convergent, or divergent? Solution: The series is absolutely convergent. To justify this, we must show that the series k a k , a k = ln k/ ( k 3 + ln k + 1), converges. Let b k = 1 /k 2 , and observe that 0 < a k < b k for all k 1. The series k b k is a p -series with p = 2 > 1; hence it converges. It follows from the basic comparison test that k a k also converges. (3) Evaluate the integral Z +1 - 1 2 x 1 - x 2 dx or explain why you can’t. Solution: This improper integral converges (and therefore can be evaluated) if and only if the improper inte- grals Z 1 0 2 x 1 - x 2 dx and Z 0 - 1 2 x 1 - x 2 dx both converge. But (by definition), Z 1 0 2 x 1 - x 2 dx = lim b 1 - Z b 0 2 x 1 - x 2 dx = lim b 1 - - ln(1 - b 2 ), which diverges. Hence, Z +1 - 1 2 x 1 - x 2 dx diverges.
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