Posted by **amber** on Monday, August 1, 2011 at 7:16pm.

Recall that a function G(x) has the limit L as x tends to infinity, written

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. for the interval 2pi to infinity.

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

with all possible L values.]

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