In class we proved that $$ \lim_{x \to \infty} \frac{x!}{2^{x}} = \infty$$
This got me thinking for what value $n$ $$ \lim_{x \to \infty} \frac{x!}{n^{x}}$$ would the limit be $= 0$.
So clearly $n = x$ makes the bottom part of the fraction go to infinity much faster than the top part, and this is the case for $n = \frac{x}{2}$ as well. However, the limit for $n = \frac{x}{3}$ is $\infty$. I immediately became suspicious that the "turning point" would be for $n = \frac{x}{e}$. Due to calculator approximation errors, a normal TI-89 says the limit is $\infty$, but I'm not really sure if that's correct.
In any case, how would one compute the limit for when $n = \frac{x}{e}$?
Answer
Hint
When you have to manipulate factorials, Stirling approximation is very often the trick to be used.
As a first approximation, you have $$x!\approx \sqrt{2\pi\, x}\,\Big(\frac x e\Big)^x$$ which than makes for your problem $$\frac{x!}{\left( \frac{x}{e} \right)^{x}}\approx \sqrt{2\pi\,x} $$ Please, remember it : it is very useful and you will often need it !
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