calculate limit

Question

$\lim_{x\rightarrow 0}\frac{1^{x}+2^{x}+3^{x}+ \dots +n^{x}-n}{x}$

Answer 1

if we put limit value in given equation then it gives $\frac{n-n}{0}= \frac{0}{0}$ which is indeterminate form so we apply l'hopital's rule which gives

$\lim_{ x \to 0} \frac{ 1^x\ln 1+2^x\ln 2 + \dots +n^x \ln n-0}{1} \\= \ln (1.2.3.......n) \\= \ln (n!)$