Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. 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

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


This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.
and here: List of abstract duplicates.

The simplest way would be; let
$$ 
\color{fuchsia}{P_n=\frac{x^n}{n!}=}
\color{maroon}{\frac x1.\frac x2.\frac x3\cdots\frac x{x-1}.\frac xx.\frac x{x+1}\cdots\frac x{n-1}.\frac xn}$$
Then
$$\color{maroon}{0}\color{red}{<}\color{fuchsia}{P_n}\color{red}{<}\color{maroon}{\frac x1.\frac x2\cdots\frac{x}{x-1}.\frac xx.}\color{green}{\frac x{x+1}.\frac x{x+1}\cdots\frac{x}{x+1}.\frac x{x+1}}$$
Or
$$\color{maroon}{0}\color{red}{<}\color{fuchsia}{P_n}\color{red}{<}\color{maroon}{\frac{x^x}{x!}.}\color{green}{\left(\frac x{x+1}\right)^{n-x}}$$
And as
$$\color{fuchsia}{\lim_{n\to\infty}\color{maroon}{0}=0}\\
\color{fuchsia}{\lim_{n\to\infty}\color{maroon}{\frac{x^x}{x!}.}\color{green}{\left(\frac x{x+1}\right)^{n-x}}=0}$$
By using $\color{red}{\text{Sandwich theorem}}$ the result can be obtained; I leave you to read between the lines.
