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Question Number 164705 by Kayela last updated on 20/Jan/22

$$\underset{x\to 0}{\lim }{\left(\frac{1^x+2^x+...+n^x}{n}\right)}^{\frac{1}{x}}$$

Answered by Berlindo last updated on 20/Jan/22

$$={\left[\underset{x\to 0}{\lim }{(1+\frac{1^{\mathrm{x}}+2^{\mathrm{x}}+...+{\mathrm{n}}^{\mathrm{x}}-\mathrm{n}}{\mathrm{n}})}^{\frac{\mathrm{n}}{1^{\mathrm{x}}+2^{\mathrm{x}}+...+{\mathrm{n}}^{\mathrm{x}}-\mathrm{n}}}\right]}^{\underset{x\to 0}{\lim }\frac{1^{\mathrm{x}}+2^{\mathrm{x}}+...+{\mathrm{n}}^{\mathrm{x}}-\mathrm{n}}{\mathrm{x}}\cdot \frac{1}{\mathrm{n}}}$$$$={\mathrm{e}}^{\underset{x\to 0}{\lim }\frac{(1^{\mathrm{x}}-1)+(2^{\mathrm{x}}-1)+...+({\mathrm{n}}^{\mathrm{x}}-1)}{\mathrm{x}}\cdot \frac{1}{\mathrm{n}}}$$$$={\mathrm{e}}^{[\underset{x\to 0}{\lim }\frac{1^{\mathrm{x}}-1}{\mathrm{x}}+\underset{x\to 0}{\lim }\frac{2^{\mathrm{x}}-1}{\mathrm{x}}+...+\underset{x\to 0}{\lim }\frac{{\mathrm{n}}^{\mathrm{x}}-1}{\mathrm{x}}]\cdot \frac{1}{\mathrm{n}}}$$$$={\mathrm{e}}^{(\ln 1+\ln 2+...+\ln \mathrm{n})\cdot \frac{1}{\mathrm{n}}}={\mathrm{e}}^{\frac{1}{\mathrm{n}}\ln (\mathrm{n}!)}={\mathrm{e}}^{\ln \left(\sqrt[\mathrm{n}]{\mathrm{n}!}\right)}$$$$=\sqrt[\mathrm{n}]{\mathrm{n}!}$$$$\therefore \underset{x\to 0}{\lim }{\left(\frac{1^x+2^x+...+n^x}{n}\right)}^{\frac{1}{x}}=\sqrt[\mathrm{n}]{\mathrm{n}!}$$$$$$$$★\mathrm{Berlindo}\mathrm{Ndala}★$$

Answered by puissant last updated on 21/Jan/22

$$=\underset{x\to 0}{\lim }{\left(\frac{e^{xln1}+e^{xln2}+...+e^{xlnn}}{n}\right)}^{\frac{1}{x}}$$$$=\underset{x\to 0}{\lim }{\left(\frac{(1+xln1)+(1+xln2)+...+(1+xlnn)}{n}\right)}^{\frac{1}{x}}$$$$=\underset{x\to 0}{\lim }{(1+\frac{xln(n!)}{n})}^{\frac{1}{x}}$$$$=\underset{x\to 0}{\lim }{e}^{\frac{1}{x}ln(1+\frac{xln(n!)}{n})}=\underset{x\to 0}{\lim }{e}^{\frac{1}{x}\left(\frac{xln(n!)}{n}\right)}$$$$=e^{\frac{ln(n!)}{n}}=\sqrt[n]{n!}..$$$$............\mathcal{L}epuissant...........$$

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