Compute $\lim_{x\to\infty} \frac{{(x!)}^{\frac{1}{x}-1} (x\Gamma(x+1) \psi^{(0)}(x+1)-x! \log(x!))}{x^2}$ What's the strategy one may use when facing a limit like this one? I think it's more important to know the possible ways to go than the answer itself. It's a problem that came to my mind again when I was working on a different problem.
$$\lim_{x\to\infty} \frac{{(x!)}^{\frac{1}{x}-1} (x\Gamma(x+1) \psi^{(0)}(x+1)-x! \log(x!))}{x^2}$$
Any suggestion, hint are very welcome.
 A: Note that
$$
\frac{\left(\Gamma(x+1)^{1/x}\right)'}{(x)'}=
\lim_{x\to\infty} \frac{{(\Gamma(x))}^{\frac{1}{x}-1} (x\Gamma(x+1) \psi^{(0)}(x+1)-\Gamma(x) \log(\Gamma(x)))}{x^2}
$$
So recalling L'Hopital's rule we see that it is enough to find 
$$
\lim\limits_{x\to\infty}\frac{\Gamma(x+1)^{1/x}}{x}
$$
We know the following asymptotic 
$$
\Gamma(x+1)\sim\left(\frac{x}{e}\right)^x\sqrt{2\pi x}\quad\text{ when }\quad x\to\infty
$$
then
$$
\lim\limits_{x\to\infty}\frac{\Gamma(x+1)^{1/x}}{x}=
\lim\limits_{x\to\infty}\frac{\frac{x}{e}(2\pi x)^{1/(2x)}}{x}=
\lim\limits_{x\to\infty}\frac{1}{e}(2\pi x)^{1/(2x)}=\frac{1}{e}
$$
As for the another one approach to the limit
$$
\lim\limits_{x\to\infty}\frac{(x!)^{1/x}}{x}
$$
see this question.
A: The numerator presents the most difficult problem. First, we can see that in this case we can simplify: $\Gamma(x+1) = x!$, so we can pull out a factor $x!$ to get
$(x!)^{1/x}(x \psi^{(0)}(x + 1) - \log(x!))$
(for the numerator.)
This is much more manageable.
Then what I'd do next is to replace the nasty functions ($x!$, $\log(x!)$ and $\psi$) with their elementary asymptotics toward infinity (Stirling's approximation and the digamma's log asymptotic). This should help.
