# Prove by definition of limit

$\displaystyle\lim_{n\to\infty} \dfrac{1}{n!} = 0$

I have no idea how to proceed with this. Usually I start with a preliminary computation and solve for $n$ in terms of $\epsilon$ from the definition of the limit:

$|\dfrac{1}{n!} - 0| < \epsilon$

Here, I do not see an approach to solve for n.

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Solving for $n$ is totally unnecessary. You are not asked to find the smallest $N$ such that beyond it $|a_n|\lt \epsilon$. You are only asked to show there is an $N$ such that beyond it, $\dots$. –  André Nicolas Sep 29 '12 at 23:00

HINT: $$\frac1{n!}\le\frac1n$$ for $n>0$. Now use the Archimedean property of the reals.
You can either try to find a $n$ for which $$n!\geq \frac{1}{\epsilon}$$ or as Brian suggested, you only need to find $n$ such that $$n\geq \frac{1}{\epsilon}.$$ In either case, it proves that the limit goes to $0$.