Dtjytyk

Consider the following expression.

$$limlimits_{xto 0} left (frac{1^x+2^x+3^x+dots+n^x}{n} right )^{frac 1 x}$$

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 $$ln(y)=lim_{xto 0}frac{1}{x}ln left (frac {1^x+2^x+cdots +n^x} {n} right ).$$

Now applying the LH rule, we get $$ln(y)=lim_{xto 0}frac{n}{1^x+2^x+cdots +n^x}({1^xln(1)+cdots +n^xln(n)})$$

Is this a right way to go?

Define $S={k mid kin mathbb{Z}^+ land kle n}$. 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.

$$begin{aligned}lim_{xto 0}left(dfrac{displaystylesum_{kin S}k^x}{n}right)^{1/x}&=lim_{xto 0} exp dfrac{1}{x}lnleft(dfrac{displaystylesum_{kin S}k^x}{n}right)\&=lim_{xto 0}exp dfrac{n}{displaystylesum_{kin S}k^x}cdotdfrac{mathrm d}{mathrm dx}left[dfrac{displaystyle sum_{kin S} k^x}{n}right]\&=lim_{xto 0}exp dfrac{n}{displaystyle sum_{kin S}k^x}cdotdfrac{displaystylesum_{kin S}k^xln k}{n}\ &=lim_{xto 0}exp dfrac{displaystylesum_{kin S}k^xln k}{displaystyle sum_{kin S}k^x}to exp dfrac{ln n!}{n}=sqrt[n]{n!}end{aligned}$$

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.

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

$$frac{n}{1^x+2^x+cdots +n^x}cdotfrac{1^xln(1)+cdots +n^xln(n)}{n}.$$

I write this answer only because the wrong one is accepted.

There is an error in your final expression. It should be:

$$lim_{xto 0}frac{n}{1^x+2^x+cdots +n^x}frac{1^xln(1)+cdots +n^xln(n)}{color{red}n}=frac{ln n!}{n}.$$

Correspondingly:
$$
limlimits_{xto 0} left (frac{1^x+2^x+3^x+dots+n^x}{n} right )^{frac 1 x}=sqrt[n]{n!}.
$$