Prove that $\lim\limits_{n \rightarrow \infty}\frac{x^n}{n!} = 0$ So I have to prove 2 things:


*

*That $\lim\limits_{n \rightarrow \infty}\frac{x^n}{n!} = 0$ where $n \in \mathbb N$ and $x \in \mathbb R, x>0$. 

*That $\lim\limits_{n \rightarrow \infty}\frac{x^n}{n!} = 0$ where $n \in \mathbb N$ and $x \in \mathbb R$. 
For #1, I know that $\frac{x^n}{n!} >0$, which means that I can find an upper bound and use squeeze theorem. For #2, I have no idea where to start.
 A: I think I've answered this question before, but note that $e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$, so for every fixed $x\in \mathbb{R}$ the series converges, thus the sequence of the terms $\{\frac{x^n}{n!}\}_{n\in \mathbb{N}}$ must converge to zero as $n$ goes to infinity, otherwise, the sum would not converge.
A: Hint: Given $x$, take $N \in \mathbb{N}$ such that $N>x$. Then, for $n >N$,
$$\frac{x^n}{n!} = \frac{x^N}{N!} \cdot \frac{x^{n-N}}{(n!/N!)} \le \frac{x^N}{N!} \cdot \left(\frac{x}{N+1}\right)^{n-N} .$$
A: A little bit of support for Koto:
For every $x \in \mathbb{R}$ the exponential series 
$\star)$ $\exp(x): = \sum_{n=0}^{\infty}\dfrac{x^n}{n!}$ converges absolutely.
Hence: $\lim_{n \rightarrow \infty} |\dfrac{x^n}{n!}| = 0$ for $x \in \mathbb{R}$.
Proof of $\star$):
Ratio test: 
With $a_n: = \dfrac{x^n}{n!}$ we get for 
$x\ne 0$ and $n\ge 2|x|$:
$|\dfrac{a_{n+1}}{a_n}| =$
$ |\dfrac{x^{n+1}}{(n+1)!}\dfrac{n!}{x^n}|=$
$\dfrac{|x|}{n+1}\le 1/2$. 
A: First Answer
The series 
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
converges then $$\frac{x^n}{n!}\to 0$$
OR
Second Answer
Use the following famous Stirling formula: Given $x>0$
$$ \lim_{n\to +\infty} \frac{n!}{\left(\frac{n}{e}\right)^n\sqrt{2n} }=\sqrt{\pi}. $$
and $$|x^n| =e^{n\ln |x|}$$
A: Note that $n!\ge (n/2)^{n/2}$.  Then, we have
$$\left|\frac{x^n}{n!}\right|\le \frac{|x|^n}{\left(\sqrt{n/2}\right)^n}=\left(\frac{2|x|}{n}\right)^n$$
Can you conclude now?
A: $$
\frac {x^n}{x!} = \frac{\overbrace{x\cdot x\cdot x\cdots\cdots x}^{\Large n \text{ factors}}}{\underbrace{1\cdot2\cdot3\cdots\cdots n}_{\Large n \text{ factors}}}
$$
When $n$ gets to be more than twice as big as $x,$ then multiplying by $\dfrac x n$ reduces the thing to less than half what it was. Then $n$ keeps growing (while $x$ does not) and at some point $n$ is more than three times as big as $x$ and then you're multiplying by $\dfrac x n < \dfrac 1 3$ and making the thing at each step less than $1/3$ what it had been. And so on$\,\ldots$ So it approaches $0.$
(That assumes $x$ is positive. If $x$ is negative, then think about $\dfrac{|x|^n}{n!}.$ It that approaches $0,$ then without the absolute value that still approaches $0.$
