Find the limit of $\lim_{n\to\infty}{a^n/n!}$ $$\lim_{n\to\infty}\frac{a^n}{n!}$$
$a>0, n\in N$
If possible, a solution through the squeeze theorem. Not sure how to solve it.

Method $1$.
Consider:
$$\sum_{n=0}^{\infty} \frac{a^n}{n!}$$
Show it converges for all $a \in \mathbb{R}$ by the ratio test. Then consequently we must have your limit going to $0$ by the divergence test for series (if it didn't go to zero then the series would diverge).
Method $2$.
Consider the sequence given by:
$$a_n=\frac{a^n}{n!}$$
Then 
$$a_{n+1}=\frac{a^{n+1}}{(n+1)!}=\frac{a}{n+1}a_n$$
And for $n+1>a \implies n>a-1$ we have:
$$a_{n+1}<a_n$$
But if $a>0,n>0$ we have $a_n>0$.
$a_n$ is eventually decreasing, and is positive. Hence the sequence must have a limit.
Call this limit $L$. Because $\lim_{n \to \infty} f(n)=\lim_{n \to \infty} f(n+1)$ and we have $a_{n+1}=a_n\frac{a}{n+1}$ we must have:
$$L=L\lim_{n \to \infty} \frac{a}{n+1}=0$$
