Question Number 159405 by qaz last updated on 16/Nov/21
$$\underset{\mathrm{x}\rightarrow+\mathrm{0}} {\mathrm{lim}}\left(\frac{\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\mathrm{k}^{\mathrm{x}} }{\mathrm{2021}}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} =? \\ $$
Answered by mindispower last updated on 16/Nov/21
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\frac{{ln}\left(\frac{\Sigma{k}^{{x}} }{\mathrm{2021}}\right)}{{x}}} \\ $$$$={e}^{\frac{{ln}\left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{2021}} {\prod}}{k}\right)}{\mathrm{2021}}} =\left(\mathrm{2021}!\right)^{\frac{\mathrm{1}}{\mathrm{2021}}} \\ $$