Thanks for the reply.

Well my educated guess will be that the limit is $\displaystyle \pi$, giving me $\displaystyle \displaystyle \bigg|\frac{\sin(n)}{n}\bigg|$

Now $\displaystyle \displaystyle \bigg|\frac{\sin(n)}{n}\bigg| \leq \frac{1}{n}$ which is obvious.

In the solutions they've got this:

Let $\displaystyle \epsilon > 0$, take $\displaystyle N > \frac{1}{\epsilon}$, then for all $\displaystyle n>N$:

$\displaystyle \displaystyle \bigg|\frac{\sin(n)}{n}\bigg| \leq \frac{1}{n} < \frac{1}{N} < \epsilon $, so the sequence tends to $\displaystyle \pi$.

Just wondering how they took $\displaystyle N > \frac{1}{\epsilon}$ in the first place?

Sorry for the confusion, really not getting this whole epsilon limit business.

Thanks again

Craig