Question Number 164702 by mathls last updated on 20/Jan/22
$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left(\frac{\mathrm{1}^{{x}} +\mathrm{2}^{{x}} +\centerdot\centerdot\centerdot+{n}^{{x}} }{{n}}\right)^{\frac{\mathrm{1}}{{x}}} =? \\ $$
Answered by mahdipoor last updated on 21/Jan/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}{ln}\left(\frac{\mathrm{1}^{{x}} +…+{n}^{{x}} }{{n}}\right)\Rightarrow{Hop}\Rightarrow \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\left(\mathrm{1}^{{x}} .{ln}\mathrm{1}+…{n}^{{x}} .{lnn}\right)/{n}}{\left(\mathrm{1}^{{x}} +…+{n}^{{x}} \right)/{n}}}{\mathrm{1}}=\frac{\left({ln}\mathrm{1}+…+{lnn}\right)/{n}}{\left(\mathrm{1}+…+\mathrm{1}\right)/{n}} \\ $$$$={ln}\left({n}!\right)/{n} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}^{{x}} +…+{n}^{{x}} }{{n}}\right)^{\frac{\mathrm{1}}{{x}}} ={e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{A}} ={e}^{{ln}\left({n}!\right)/{n}} =\left({n}!\right)^{\mathrm{1}/{n}} \\ $$