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Posted by 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.]

  • calculus - , Friday, March 9, 2012 at 8:46pm

    For any divergence especially with trig definitions in infinite series you will want to make sure that you can prove it to something. Comparison test or limit comparison works really well just remember that the sine function is only good from -1 to 1

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