Why is the limit of the factorial function divided by stirling's approximation the 12th root of e?

Question

I was interested in the limit definition of $e$, where as $x$ tends towards infinity, the limit of $1$ plus $x$ to the negative first power, all to the power of $x$, is in fact $e$. I was wondering if the limit of the ratio between the factorial function and Stirling's approximation would converge in a similar manner. The formula in question and the Wolfram Alpha result are: $$ \lim_{x\to\infty} (\frac {x!}{\sqrt{2\pi x}(\frac{x}{e})^x})^x = \sqrt[12]{e} $$

Wolfram Alpha evaluated this limit as being the 12th root of $e$. My question being, why is this? I understand that this would normally be equal to $e$ itself, but what is it about the difference between this approximation and the actual factorial function that takes $e$ to the 12th root? Why 12 of all roots? Why not 13? or 14? I'm sure there is an explanation for this, I am just unsure why.

Answer 1

It comes from the next order Stirling approximant. To that order, $x!=e^{x \ln(x) - x + \frac{1}{2} \ln(2\pi x) + \frac{1}{12x}+o(1/x)}$. The denominator cancels the first three terms in the exponent, so you're left with $e^{\frac{1}{12x}}$ to leading order, and raising that to the $x$ gives the result.

This is actually related to the $(1+1/x)^x$ definition of $e$, in that the inside of the parentheses behaves to leading order like $1+1/(12x)$.