Limit of: $\lim\limits_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}$ Compute the following limit ($n\in \mathbb{N}$)
$$\lim_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}$$
My idea was to use the inequality:
$$\left(\frac{1}{n}\sum_{k=1}^{n} 1^{1/x}\right)^{nx}<\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}<\left(\frac{1}{n}\sum_{k=1}^{n} n^{1/x}\right)^{nx} \\
\implies1<L<n^n$$
This gives that the required limit $L$ lies between $1$ and $n^n$. But how can we find its value?
 A: Since $e^t=1+t+O\!\left(t^2\right)$, set $t=\frac{\log(k)}x$
$$
\begin{align}
\lim_{x\to\infty}\left(\frac1n\sum_{k=1}^nk^{1/x}\right)^{nx}
&=\lim_{x\to\infty}\left(1+\frac1n\sum_{k=1}^n\frac{\log(k)}x+O\!\left(\frac1{x^2}\right)\right)^{nx}\\
&=\lim_{x\to\infty}\left(1+\frac1n\sum_{k=1}^n\frac{\log(k)}x\right)^{nx}\lim_{x\to\infty}\left(1+O\!\left(\frac1{x^2}\right)\right)^{nx}\\[3pt]
&=\lim_{x\to\infty}\left(1+\frac1n\frac{\log(n!)}x\right)^{nx}\cdot1\\[9pt]
&=e^{\log(n!)}\\[15pt]
&=n!
\end{align}
$$
A: Let  $$f(h)= \ln\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h}\right)\implies f'(h) =\frac{\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h}\ln k \right)}{\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h} \right)}$$
$$f(0)= 0 ~~~and~~~~f'(0)=\left(\frac{1}{n}\sum\limits_{k=1}^{n} \ln k\right)=\color{blue}{ \frac{1}{n}\ln \left(n!\right) } $$ 
Let $x=1/h$ then we have
$$\lim_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx} =\lim_{h\to 0}\exp\left(\frac{n}{h}\ln\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h}\right)\right)\\ =\lim_{h\to 0}\exp\left(n\frac{f(h)}{h}\right)
   =\color{red}{\exp\left(nf'(0)\right) }=\color{red}{n! }  $$
