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Unformatted text preview: dx x 2 (1+ e x ) is discontinuous at x = 0 and so two limits are required for 0and 0 + ; and 2) the interval of the integrations includes and , and so two limits are required for and . Then lim x 1 x 2 1 x 2 (1+ e x ) = lim x x 2 (1+ e x ) x 2 = lim x (1 + e x ) = 1 2 and R 1 dx x 2 diverges R 1 dx x 2 (1+ e x ) diverges by another version of Limit Comparison Test that I mentioned in section R  dx x 2 (1+ e x ) diverges. 1...
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 Spring '07
 BERMAN
 Math, Calculus

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