How to find limit of the given problem?

Question

How to find this limit ? I've tried this problem but in the middle steps i'm stuck. I'll upload the image of the solution

Answer 1

The correct explanation of my Question is this...

Answer 2

It's $$\lim_{x\rightarrow0}\left(1+\frac{1^x+2^x+...+n^x}{n}-1\right)^{\frac{n}{1^x-1+2^x-2+...+n^x-n}\cdot\frac{\frac{2^x-1}{x}+...+\frac{n^x-1}{x}}{n}}=e^{\frac{\ln2+...+\ln n}{n}}=\sqrt[n]{n!}.$$ I used the following.

1.For all $a>0$ we have $$\lim_{x\rightarrow0}\frac{a^x-1}{x}=\ln{a}.$$

2. Since $\frac{1^x+2^x+...+n^x}{n}-1\rightarrow0$ for $x\rightarrow0$ and $\lim\limits_{x\rightarrow0}(1+x)^{\frac{1}{x}}=e$, we obtain $$\lim_{x\rightarrow0}\left(1+\frac{1^x+2^x+...+n^x}{n}-1\right)^{\frac{n}{1^x-1+2^x-2+...+n^x-n}}=e$$
3. If there is $\lim\limits_{x\rightarrow a}f(x)=A,$ where $A>0$ and there is $\lim\limits_{x\rightarrow a}g(x)=B$ then there is $$\lim_{x\rightarrow a}f(x)^{g(x)}=A^B.$$