Proving that if $a_n\in\mathbb{Z}$ for all $n$ then it's limit is also in $\mathbb{Z}$?

Edit: Well, this is awkward, after asking the question (that did frustrate me for a few days) I was able to solve it myself without using (and even before looking at) the solutions provided. What am I supposed to do now?

Let $\{a_n\}_{n=1}^\infty$ a sequence converging to $L$. Prove that: c) If $\forall n\in\mathbb{N}$,$a_n\in\mathbb{Z}$ then $L\in\mathbb{Z}$

What I think I should do is: Assume for contradiction that $L\notin\mathbb{Z}$, Therefore exists $a\in\mathbb{Z}$ such that $a-1<L<a$. And since the sequence is converging I can choose an epsilon such that:

$L-\epsilon<a_n<L+\epsilon$

Therefore I need to find an epsilon such that:

$a-1\leq L-\epsilon<a_n<L+\epsilon\leq a$

But I really have no idea what to do here, I can only think of epsilons that are true for one side ($\epsilon=L-a+1$ or $\epsilon=a-L$)

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In response to your edit: well done! The purpose of this site is it provide help with maths problems, and if you've found out how to do it on your own then that's even better than doing it with our hints (and much better than just taking one of our answers!) The usual protocol is to write an answer yourself, and accept that one. It seems slightly unusual I know, but it's an allowed and encouraged thing to do (see the faq). – Tom Oldfield Nov 25 '12 at 13:17

If $L\notin \mathbb Z$ then $\exists \varepsilon > 0 : |L-x|<\varepsilon \Rightarrow x \notin \mathbb Z$.
Hint: prove that, under the given data, the sequence $\,\{a_n\}\,\subset \Bbb Z\,$ is eventually constant, meaning:
$$\{a_n\}\subset\Bbb Z\,\,,\,\,\lim_{n\to\infty}a_n=L\Longleftrightarrow\,\exists \,N\in\Bbb N\,\,s.t.\,\,n>N\Longrightarrow a_n=L$$