calculus
posted by Anna .
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.]

The integral of sinx is from 2pi to infinity