Limit of a function with factorial

Question

$\lim_{n \rightarrow \infty} (x^n/n!)=0$. prove. x is finite whereas n is infinite. But increasing n means also increasing $x^n$. It is understandable that if n is too large n! will exceed $x^n$. How it can be proved in a mathematically precise way?

Answer 1

Suppose that $0\lt |x|\le N$, where $N$ is a positive integer, and let $n=2N+k$. Then $$\frac{|x|^n}{n!}=\frac{|x|^{2N}}{(2N)!}\frac{|x|}{2N+1}\cdot \frac{|x|}{2N+2}\cdot\frac{|x|}{2N+3}\cdots \frac{|x|}{2N+k}.\tag{1}$$ Each of the terms on the right-hand side of (1) except the first is $\lt \frac{1}{2}$. It follows that $$\frac{|x|^n}{n!}\lt \frac{|x|^{2N}}{(2N)!}\frac{1}{2^k}.\tag{2}$$ As $k\to\infty$, the expression on the right-hand side of (2) approaches $0$.

Remark: In the calculation above, $x$ is fixed. So $\frac{|x|^{2N}}{(2N)!}$ is fixed, and possibly quite large. But after that, each time we increment $n$ by $1$, the value of $\frac{|x|^n}{n!}$ decreases by a factor of at least $2$.