# Factorial limit

• Jan 20th 2013, 01:05 PM
Factorial limit
How do we show that

$\displaystyle \lim_{x \rightarrow \infty} \, \frac {e^x (x-1)!}{x!} = + \infty \, .$

I know that the factorial functionen approaches infinity faster than most other elementary functions e.g. $\displaystyle e^x$. But I am not sure how to think when we have a factorial function in both the nominator and denominator.
• Jan 20th 2013, 02:13 PM
tom@ballooncalculus
Re: Factorial limit
$\displaystyle \frac{(x - 1)!}{x!}\ =\ \text{... ?}$
• Jan 20th 2013, 02:30 PM
hollywood
Re: Factorial limit
I would start by cancelling. Since $\displaystyle x! = x(x-1)(x-2)\dots 1$ and $\displaystyle (x-1)! = (x-1)(x-2)\dots 1$, $\displaystyle \frac{x!}{(x-1)!}=x$ and so $\displaystyle \lim_{x \rightarrow \infty} \, \frac {e^x (x-1)!}{x!} = \lim_{x \rightarrow \infty} \, \frac {e^x}{x}$. And now I'll bet it looks like problems you've done before.

- Hollywood

P.S. I have to start writing faster - tom@ballooncalculus beat me to the answer.
• Jan 20th 2013, 02:34 PM
$\displaystyle \frac{(x - 1)!}{x!}\ =\ \text{... ?}$