Question

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.

[Hint: Consider the cases L >(or equal t0) 1 and L < 1 in order to deal
with all possible L values.]

Answer 1

The integral of sinx is from 2pi to infinity