Solving limits involving factorials

Question

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?

Answer 1

$x^n$ is a $o( n! )$ (little o) which means that exponentials grow much slower than factorials.

https://en.wikipedia.org/wiki/Big_O_notation

Answer 2

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 $$a_n= \frac{|x|^{(n+1)}}{(n+1)!}.$$

Answer 3

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

$\exists\;\lim\limits_{n\to\infty}a_n=0\;.$