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

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, $\dots$.2012-09-29

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HINT: $\frac1{n!}\le\frac1n$ for $n>0$. Now use the Archimedean property of the reals.

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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 $0$.