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

First you show that $n!>3^n$ and then use $$ \lim\limits_{n}\frac{2^n}{n!}\leq \lim\limits_n\frac{2^n}{3^n} =\lim\limits_n\left(\frac2{3}\right)^n = 0. $$

To show that $n!>3^n$ you use induction. For $n = 7$ it holds, you assume that it holds for some $k\geq7$ then $(k+1)! = k\cdot k!>k\cdot 3^k>3^{k+1}$ since $k\geq 7>3$.

Consider that $$\frac{2^n}{n!} = \frac{\overbrace{2\times 2\times\cdots \times 2}^{n\text{ factors}}}{1\times 2 \times \cdots \times n} = \frac{2}{1}\times \frac{2}{2}\times \frac{2}{3}\times\cdots \times\frac{2}{n}.$$ Every factor except the first two is smaller than $1$, so at each step you are multiplying by smaller and smaller numbers, with the factors going to $0$.