Why does $\lim\limits_{x \to \infty}{\frac{a^x}{x!}}\;,\;a \in \mathbb{R}^+,$ approach $0?$ I'm currently writing a paper about numerical analysis and at one point I needed to calculate  $$\lim\limits_{x \to \infty}{\frac{a^x}{x!}}\quad \text{with} \quad a \in \mathbb{R}^+$$  Based on the fact that $x!$ goes faster to infinity than $a^x,$ I thought that it would approach zero, but this isn't really a proof, only a thought.
So I'd appreciate it if you'd be able to help me with proving this.
 A: This is a fairly standard lemma in undergraduate calculus courses, to the point that in my opinion in a paper you definitely do not need to include a proof, or even a reference.
So... basically: if $0<a \leq 1$, then $\frac{a^n}{n!} \leq \frac{1}{n!}$ and it's obvious.
If $a > 1$, apply the ratio test: $\frac{a^{n+1}}{(n+1)!}\cdot \frac{n!}{a^n} = \frac{a}{n+1}$ which tends to $0$.
Not a proof, but an intuitive explanation of the fact that, asymptotically, $a^n < n!$ can be obtained by taking the $\log$ of both sides:
$$n \log(a) <^? \sum_{i=1}^n \log(i)$$
from which you can see that, as $n$ increases to $n+1$, you always add $\log(a)$ to the left (constant increase) but you add $\log(n+1)$ to the right, which is bigger for $n$ big enough. It stands to reason that then, eventually, the right side will prevail.
A: We can think the integer case as following:
$$
\frac {a^n}{n!} = \frac{a\cdot a\cdot a\cdots a}{n\cdot (n-1)\cdot (n-2)\cdots 1}
$$
when $n\to \infty$, especially when $n>N$ where $ N\le a < N+1$ we can observe that
$$
\frac {a^n}{n!} = \frac{a\cdot a\cdot a\cdots a}{n\cdot (n-1)\cdot (n-2)\cdots 1}=\frac{a\cdot a\cdot a\cdots a}{(N+1)\cdot (N)\cdot (N-2)\cdots 1} \frac{a\cdot a\cdot a\cdots a}{n\cdot (n-1)\cdot (n-2)\cdots (N+2)}
$$
Note that one part is constant , and one part is Infinitesimal
A: *

*Tool:  Stirling's approximation

$$\begin{align}\color {gold}{\boxed {\color{black}{\sqrt{2\pi}\ n^{n+\frac12}e^{-n} \le n! \le e\ n^{n+\frac12}e^{-n}.}}}\end{align}$$
$$\begin{align}0≤{\frac{a^x}{x!}}≤\frac{a^x}{x^xe^{-x}}=\frac {a^x e^x}{x^x}=\frac{(ae)^x}{x^x}\end{align}$$
We have,
$$\begin{align}0≤\lim\limits_{x \to \infty}\frac {(ae)^x}{x^x}&=\lim\limits_{x \to \infty}\left(\frac{ae}{x}\right)^x\\ &≤\lim\limits_{x \to \infty}\left(\frac{\sqrt x}{x}\right)^x\\ &= \lim\limits_{x \to \infty}\left(\frac{1}{\sqrt x}\right)^x \\ &=0.\end{align}$$
Finally,  apply the Squeeze theorem.
A: $$y=\frac{a^x}{x!}\implies \log(y)=x \log(a)-\log(x!)$$ Using Stirling approximation
$$\log(y)=x (\log (a)-\log (x)+1)-\frac{1}{2} \log (2 \pi  x)+O\left(\frac{1}{x}\right)$$ This already shows that $\log(y)\to -\infty$ which implies that $y \to 0$. If you want more
$$y=e^{\log(y)}=\frac{1}{\sqrt{2 \pi x}} \left(\frac{e a}{x}\right)^x+O\left(\frac{1}{x}\right)$$
