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.


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 $k\geq 7>3$.

Source : Link , Question Author : Matt Nashra , Answer Author : Ilya

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