Solving limits involving factorials Generally speaking, how does one handle a limit involving a factorial? Take, for example, the limit:
$$\lim_{n\to\infty} \frac{x^{n+1}}{(n+1)!}$$
According to WolframAlpha, this limit evaluates to $0$, which seems reasonable. How can this be determined?
 A: Hint :
If $$ \{a_n\} $$ is a sequence of positive real numbers and $$\lim_{n \to \infty} \frac {a_{n+1}}{a_n} = l. $$ where $$ 0 \lt l \lt 1$$ then $$ \lim a_n = 0$$.
Here $$a_n= \frac{|x|^{(n+1)}}{(n+1)!}.$$
A: Let $\;a_n=\dfrac{x^{n+1}}{(n+1)!}\quad$ for all $\;n\in\mathbb{N}\;.$
It results that
$a_{n+1}=\dfrac{x^{n+2}}{(n+2)!}=\dfrac{x}{n+2}\cdot\dfrac{x^{n+1}}{(n+1)!}=\dfrac{x}{n+2}\cdot a_n$
for all $\;n\in\mathbb{N}\;,$
hence,
$|a_{n+1}|=\dfrac{|x|}{n+2}\cdot |a_n|\le|a_n|$
for all $\;n\in\mathbb{N}\land n\ge |x|-2\;,$
consequently the sequence $\;\big\{|a_n|\big\}_{n\in\mathbb{N}}\;$ is eventually decreasing and
$\exists\;\lim\limits_{n\to\infty}|a_n|=\inf\limits_{n\in\mathbb{N}\\n\ge|x|-2}\big\{|a_n|\big\}=l\ge0\;.$
Since $\;|a_{n+1}|=\dfrac{|x|}{n+2}\cdot |a_n|\;\;$ for all $\;n\in\mathbb{N}\;,\;$ then
$\lim\limits_{n\to\infty}|a_{n+1}|=\lim\limits_{n\to\infty}\dfrac{|x|}{n+2}\cdot\lim\limits_{n\to\infty}|a_n|\;,$
$l=0\cdot l\;,$
$l=0\;,\;$ therefore
$\exists\;\lim\limits_{n\to\infty}|a_n|=0\;,\;$ hence
$\exists\;\lim\limits_{n\to\infty}a_n=0\;.$
A: $x^n$ is a $o( n! )$ (little o) which means that exponentials grow much slower than factorials.
https://en.wikipedia.org/wiki/Big_O_notation
