Pre-Calc Homework Solutions 359

4 dx x 2 16 lim tan b t 1 4t dt t 2t 1t 2 1 dt t2 dt

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Unformatted text preview: dt t 1 2t t 2 4 2 1 dt 1 1 dt t2 1 2 2 2 1 2 2t dt t2 1 49. Use the limit comparison test with f( ) g( ) 1 f( ) g( ) 1 and 1 . Both are positive continuous functions on [1, ). 2 2 ln t ln 3 1 2 ln t tan 1 t C tan 1 t C lim lim 1 1 1 lim 1 1 2 1 (t 1) t2 1 1 Since g( ) d d b t 1 4t dt t 2(t 1)(t 2 1) 1 lim 0 1 1 1/2 1 t 1 4t dt t 2(t 1)(t 2 1) 4t 3 t 1 dt t (t 1)(t 2 1) 2 3 3 t 1 4t dt t 2(t 1)(t 2 1) 4t 3 t 1 dt t (t 1)(t 2 1) 2 3 3 b 1 d b lim ln b 1 1/2 0 2 1 lim ln b b t 1 4t dt t 2(t 1)(t 2 1) 2 t 1 4t dt t 2(t 1)(t 2 1) , we know that 1 g( ) d diverges and so 1 f( ) d diverges. This means that the given integral diverges....
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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