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

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


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$u_n=\dfrac{x^n}{n!} \implies \dfrac{u_{n+1}}{u_n}=\dfrac{x^{n+1}n!}{x^n(n+1)!}=\dfrac{x}{n+1}$
$\therefore\displaystyle\lim_{n \to \infty}\dfrac{u_{n+1}}{u_n}=0$
$\therefore\displaystyle\lim_{n \to \infty}u_n=0$

Since for a sequence $\{u_n\}$ of positive real numbers such that $\displaystyle\lim_{n \to \infty}\dfrac{u_{n+1}}{u_n}=L\ (<1)$ we must have $\displaystyle\lim_{n \to \infty}u_n=0$.

