Proving Limit of sin (n!*R*π) for Rational R

[happyg1]

Hi,
Here is my dilemma: I am to prove that sin (n!*R*pi) has a limit, where R is a rational number. I rewrite R as a/b and I can see that whenever n=b, every subsequent term will be zero. I have tried to write this out using the definition of a limit, but I can't seem to break it down. I have been looking at these problems for a long time and I am blocked on this one.
Thanks in advance,
CC

 

[Physics Monkey]

For a sequence a_n, we say that a_n \rightarrow L if for every \epsilon > 0 we can find an N(\epsilon) such that | a_n - L | < \epsilon for n > N(\epsilon).

You have guessed correctly that the limit is 0. Now I give you an \epsilon and ask you to tell me where I should start looking so that the terms of the sequence are always closer than \epsilon to the limit. Tell me where to look by giving me N. You've already pointed out that the terms of the sequence equal the limit beyond a certain point ...

 

[happyg1]

hey,
I think I got it. I wasn't including my b in the expansion of the n! as the spot where the sequence converges. I couln't relate the epsilon to the b or the n. It's just the algebra. That's what was giving me the headache. I had been doing the ones that are all polynomials and my brain was fried.

CC