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November 24, 2014

November 24, 2014

Posted by **Anna** on Sunday, July 31, 2011 at 6:41pm.

lim(x->infinity)G(x) = L,

if for any epsilon > 0, there exists M >0 so that if x > M, then |G(x) − L| < epsilon.

This means that the limit of G(x) as x tends to infinity does not exist if for

any L and positive M, there exists epsilon > 0 so that for some x > M,

|G(x) − L| >(or equal to) epsilon.

Using this definition, prove that

the indefinite integral of sin(theta)

diverges.

[Hint: Consider the cases L >(or equal t0) 1 and L < 1 in order to deal

with all possible L values.]

- calculus -
**Anna**, Sunday, July 31, 2011 at 6:43pmThe integral of sinx is from 2pi to infinity

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