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Question Number 75939 by ahmadshahhimat775@gmail.com last updated on 21/Dec/19

Commented by JDamian last updated on 21/Dec/19

$$c){(n!)}^{\frac{1}{n}}$$

Commented by mathmax by abdo last updated on 21/Dec/19

$$letA\left(x\right)={\left(\frac{1^x+2^x+3^x+...+n^x}{n}\right)}^{\frac{1}{x}}\Rightarrow$$$$A\left(x\right)=e^{\frac{1}{x}ln\left(\frac{1^x+2^x+....+n^x}{n}\right)}letfind{lim}_{x\to 0}\frac{1}{x}ln\left(\frac{1^x+2^x+3^x+...+n^x}{n}\right)$$$$letusehospitaltheoremwithu\left(x\right)=ln(1^x+2^x+3^x+...+n^x)$$$$andv=x$$$$wehave{u}^{{}^{\prime }}\left(x\right)=\frac{{(1^x+2^x+3^x+...n^x)}^{{}^{\prime }}}{1^x+2^x+3^x+...+n^x}=\frac{{(1+e^{xln\left(2\right)}+e^{xln\left(3\right)}+...e^{xln\left(n\right)})}^{{}^{\prime }}}{1^x+2^x+3^x+...+n^x}$$$$=\frac{ln\left(2\right)2^x+ln\left(3\right)3^x+...+ln\left(n\right)n^x}{1^x+2^x+3^x+...+n^x}\Rightarrow$$$${lim}_{x\to 0}{u}^{{}^{\prime }}\left(x\right)=\frac{ln\left(1\right)+ln\left(2\right)+...ln\left(n\right)}{n}=\frac{ln(n!)}{n}\Rightarrow$$$${lim}_{x\to 0}A\left(x\right)=e^{\frac{ln(n!)}{n}}={\left(e^{ln(n!)}\right)}^{\frac{1}{n}}={(n!)}^{\frac{1}{n}}$$

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