How do I evaluate $\lim_{x\to 0}\left( \frac{1^x+2^x+3^x+\cdots+ n^x}{n}\right)^{a/x} $? $$\lim_{x\to 0}\left( \frac{1^x+2^x+3^x+\cdots+ n^x} n \right)^{a/x} $$ 
 A: Hint: Apply $\ln$ to get
$$a\cdot\frac{\ln (1^x + \cdots +n^x) - \ln n}{x}.$$
Let $f(x) = \ln (1^x + \cdots +n^x).$ Then the above equals
$$a\cdot\frac{f(x) - f(0)}{x-0},$$
which hopefully looks familiar.
A: we have
$$j^x=e^{x\ln (j)}=1+x\ln (j)+x\epsilon (x) .$$
then
$$S=\sum_{j=1}^nj^x=n+x\ln (n!)+x\epsilon (x).$$
and
$$\frac S n=1+\frac x n\ln (n!)+x\epsilon (x). $$
thus
$$(\frac S n)^\frac a x =e^{\frac a x \ln(1+\frac x n\ln (n!)+x\epsilon (x))}. $$
using the fact that $$\ln (1+X)\sim X \;\;(X\to 0) $$, we find the limit 
$$\boxed {e^{\frac {a}{n}\ln (n!)}= (n!)^\frac a n }$$
A: Case 1: $a\geq 0$
First; consider any decreasing positive sequence, say $(x_k)_{k=1}^{\infty}$ which decreases to $0$.
Using Jensen's inequality on concave function, we have
$$\frac{1+(2^{x_k})^{\frac{x_{k+1}}{x_k}}+(3^{x_k})^{\frac{x_{k+1}}{x_k}}+\dots +(n^{x_k})^{\frac{x_{k+1}}{x_k}}}{n}\leq \Big{(} \frac{1+2^{x_k}+3^{x_k}\dots +n^{x_k}}{n} \Big{)}^{\frac{x_{k+1}}{x_k}}$$
Hence $\Big{(} \frac{1+2^{x_k}+3^{x_k}\dots +n^{x_k}}{n} \Big{)}^{\frac{a}{x_k}}$ is a decreases as $x_k$ decreases.
Second, by AM-GM inequality,$$\Big{(} \frac{1+2^{x_k}+3^{x_k}\dots +n^{x_k}}{n} \Big{)}^{\frac{a}{x_k}}>(1\times2\times\dots\times n)^{\frac{x_k}{n}\frac{a}{x_k}}=(n!)^{\frac{a}{n}}$$.
Case 2: $a<0$
Everything will be same. Just the inequalities will reverse, then $\Big{(} \frac{1+2^{x_k}+3^{x_k}\dots +n^{x_k}}{n} \Big{)}^{\frac{a}{x_k}}$ is a increases as $x_k$ decreases.
These two cases will show that as $x\to 0^+$, the limit will be $(n!)^{\frac{a}{n}}$.
In the similar fashion you can prove that as $x\to 0^-$, the limit will be $(n!)^{\frac{a}{n}}$.
Hence, $$\lim_{x\to 0}\left( \frac{1^x+2^x+3^x+\cdots+ n^x} n \right)^{a/x}=(n!)^{\frac{a}{n}}\space \space \space \blacksquare $$
