Find $\lim\limits_{x\to 0}\left (\frac{1^x+2^x+3^x+\dots+n^x}{n} \right)^{\frac1x}$ Consider the following expression.
$$\lim\limits_{x\to 0} \left (\frac{1^x+2^x+3^x+\dots+n^x}{n} \right )^{\frac 1 x}$$
How to solve this?
Let $y= \left (\frac {1^x+2^x+\cdots +n^x} {n} \right)^{1/x}$
I tried taking  $\ln$ on both sides. We get that $$\ln(y)=\frac{1}{x}\ln \left (\frac {1^x+2^x+\cdots +n^x} {n} \right ).$$
Taking $\lim$ on both sides we get $$\ln(y)=\lim_{x\to 0}\frac{1}{x}\ln \left (\frac {1^x+2^x+\cdots +n^x} {n} \right ).$$
Now applying the LH rule, we get $$\ln(y)=\lim_{x\to 0}\frac{n}{1^x+2^x+\cdots +n^x}({1^x\ln(1)+\cdots +n^x\ln(n)})$$
Is this a right way to go?
 A: Actually, $\ln(y)=\frac{1}{x}\ln{\frac{(1^x+2^x+...+n^x)}{n}}$. Whenever after putting $x=0$ you get the form $1^{\infty}$, you need to take $\ln$ and find the limit. 
Here, after putting $x=0$, we have $\big(\frac{1^0+2^0+\cdots +n^0}{n}\big)^{1/0}=1^{\infty}$. So taking $\ln$ in both side and finding limit is correct approach.  
A: Define $S=\{k \mid k\in \mathbb{Z}^+ \land k\le n\}$. Define the limit as $L$. Rewrite it as $\exp \ln L$ and use L'Hopital's rule and you'll be done in no time.
$$\begin{aligned}\lim_{x\to 0}\left(\dfrac{\displaystyle\sum_{k\in S}k^x}{n}\right)^{1/x}&=\lim_{x\to 0} \exp \dfrac{1}{x}\ln\left(\dfrac{\displaystyle\sum_{k\in S}k^x}{n}\right)\\&=\lim_{x\to 0}\exp \dfrac{n}{\displaystyle\sum_{k\in S}k^x}\cdot\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{\displaystyle \sum_{k\in S} k^x}{n}\right]\\&=\lim_{x\to 0}\exp \dfrac{n}{\displaystyle \sum_{k\in S}k^x}\cdot\dfrac{\displaystyle\sum_{k\in S}k^x\ln k}{n}\\ &=\lim_{x\to 0}\exp \dfrac{\displaystyle\sum_{k\in S}k^x\ln k}{\displaystyle \sum_{k\in S}k^x}\to \exp \dfrac{\ln n!}{n}=\sqrt[n]{n!}\end{aligned}$$
A: Your basic approach is fine but there are some problems. The first is where you say "taking $\lim$ on both sides". You can't take the limit unless you know the limit exists. But that is part of the exercise, right? I would omit the $\lim$ business until the end. Second, why do you still have $\ln y$ on the left after "taking the limit"? Finally, you didn't get the correct expression in using LHR. It should be
$$\frac{n}{1^x+2^x+\cdots +n^x}\cdot\frac{1^x\ln(1)+\cdots +n^x\ln(n)}{n}.$$
A: I write this answer only because the wrong one is accepted.
There is an error in your final expression. It should be:
$$\lim_{x\to 0}\frac{n}{1^x+2^x+\cdots +n^x}\frac{1^x\ln(1)+\cdots +n^x\ln(n)}{\color{red}n}=\frac{\ln n!}{n}.$$
Correspondingly:
$$
\lim\limits_{x\to 0} \left (\frac{1^x+2^x+3^x+\dots+n^x}{n} \right )^{\frac 1 x}=\sqrt[n]{n!}.
$$
