How to calculate limit of a function having factorial in denominator

$$0<\frac{2^n}{n!}=\frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \dots \cdot \frac{2}{ n} \leq \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \dots \cdot \frac{2}{3}=\frac{2}{1} \cdot \frac{2}{2} \cdot \left (\frac{2}{3} \right )^{n-2}=2 \left ( \frac{2}{3} \right )^{n-2}$$

As $n \rightarrow \infty$, $\left ( \frac{2}{3} \right )^{n-2} \rightarrow 0$

Therefore, from the Squeeze Theorem $$\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0$$

BIG HINT:

$$\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}=e^x<\infty$$

You can use the theorem of d'Alembert for the sequences then you immediately have:

if $x_n=\frac{2^n}{n!}$,

$$\lim_{n\to\infty }\left|\frac{x_{n+1}}{x_n} \right|=\lim_{n\to\infty }\frac{2^{n+1}n!}{(n+1)! 2^n}=\lim_{n\to\infty }\frac{2}{(n+1)}=0<1$$

then $$\lim_{n\to\infty }\frac{2^n}{n!}=0.$$