TITLE: Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.
QUESTION [83 upvotes]: Why is

$$\lim_{n \to \infty} \frac{2^n}{n!}=0\text{ ?}$$

Can we generalize it to any exponent $x \in \Bbb R$? This is to say, is

$$\lim_{n \to \infty} \frac{x^n}{n!}=0\text{ ?}$$


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REPLY [2 votes]: Since $\frac{-|x|^n}{n!} \leq \frac{x^n}{n!} \leq \frac{|x|^n}{n!}$
Therefore, enough to show that  $\frac{|x|^{n}}{n!} \rightarrow 0$ as $n \rightarrow \infty$ 
Now $x$ is any real number, we can find $M$ such that $|x| < M$
which means $\frac{|x|}{M} < 1$, $\big(\frac{|x|}{M}\big)^n \rightarrow 0$
Now for all $n > M$ 
we have $\frac{|x|^n}{n!} = \frac{|x|^n}{1.2.3....M(M+1)...n} $ 
$\leq \frac{|x|^n}{M!M^{n-M}} $ = $\big(\frac{|x|}{M!}\big)^n \frac{M^M}{M^n} $
which will tend to $0$ as $n \rightarrow \infty$