Compute $\lim_{x\to 1^+} \lim_{n\to\infty}\frac{\ln(n!)}{n^x} $ May someone give me a hand on this double limit? Does the order of limits impact the result?
$$\lim_{x\to 1^+} \lim_{n\to\infty}\frac{\ln(n!)}{n^x} $$
I showed that the interior of the limits is inferior to the following expression: $$  \frac{\ln(n)}{{n^{x-1}}} $$
Thanks in advance :)
 A: HINT:
Note that since $\log(n!)=\sum_{k=1}^n \log(k)=\sum_{k=1}^n \log(k/n)+n\log(n)$, we have
$$\frac1{n^x}\log(n!)=\frac{1}{n^{x-1}}\underbrace{\left(\frac1n\sum_{k=1}^n \log(k/n)\right)}_{\text{Riemann Sum of}\,\int_0^1 \log(x)\,dx=-1}+\frac{\log(n)}{n^{x-1}}$$
A: As $n\to \infty,$ $\ln n! \sim n\ln n,$ hence
$$\frac{\ln n!}{n^x}\sim \frac{\ln n}{n^{x-1}}.$$
The limit of the expression on the right is $0$ if $x>1,$ and is $\infty$ if $x\le 1.$ Thus your limit as $x\to1^+$ is $0.$
A: Note that for convexity
$$0 \le \frac{\ln n!}{n^x}\leq \frac{n}{n^x}\frac{\sum_{k=1}^{n}\ln k}{n}\le\frac{1}{n^{x-1}}\ln\left({\frac{\sum_{k=1}^nk}{n}}\right)=\frac{\ln \left(\frac{n+1}2{}\right)}{n^{x-1}}\to0 \quad \forall x>1$$
thus for squeeze theorem
$$\lim_{x\to 1^+} \lim_{n\to\infty}\frac{\ln n!}{n^x}=0$$
A: With fixed $x>1$
$$(n+1)^x-n^x=\int_{n}^{n+1}xt^{x-1}dt>n^{x-1}$$
and using Stolz–Cesàro theorem
$$\lim_{n\to\infty}\frac{\ln(n!)}{n^x}=\lim_{n\to\infty}\frac{\ln(n+1)}{(n+1)^x-n^x}<\lim_{n\to\infty}\frac{\ln(n+1)}{n^{x-1}}\to0$$
then
$$\lim_{x\to 1^+} \lim_{n\to\infty}\frac{\ln(n!)}{n^x}\to0$$
