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.
 A: We know $$\mathrm{e}^x = \sum_{n=0}^{\infty} \frac{ x^n }{n!} $$
Has radius of convergence $R=\infty$. Thus, it better converge for $x=a$. It follows then that 
$$ \lim_{n \to \infty} \frac{ a^n }{n!} = 0 $$
A: Choose $N\ge 2|a|$, then for $n > N$ we have
$| { |a|^n \over n!} | \le { |a|^N \over N!}{1 \over 2^{n-N}}$, from which 
it follows that $\lim_{n \to \infty} | { |a|^n \over n!} |  = 0$.
A: 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:
$$L=L\lim_{n \to \infty} \frac{a}{n+1}=0$$
A: Consider $$A_n=\frac{a^n}{n!}\implies \log(A_n)=n \log(a)-\log(n!)$$ and use Stirling approximation for $\log(n!)$. This will give
$$\log(A_n)=n (\log (a)-\log (n)+1)-\frac{1}{2} \left(\log \left(n\right)+\log (2 \pi
   )\right)-\frac{1}{12 n}+O\left(\frac{1}{n^3}\right)$$ So, the key term is $-n\log(n)$ which makes $\log(A_n)\to-\infty$ which implies $A_n \to 0$.
When you have problems of limits for expressions containing factorials, Stirling approximation is, most of the time, the key tool to use.
