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Question Number 127961 by bramlexs22 last updated on 03/Jan/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\sqrt[\mathrm{n}]{\mathrm{a}}+\sqrt[\mathrm{n}]{\mathrm{b}}}{2}\right)}^{\mathrm{n}}=?$$$$\mathrm{a},\mathrm{b}\in \mathbb{R}$$

Answered by liberty last updated on 03/Jan/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\sqrt[\mathrm{n}]{\mathrm{a}}+\sqrt[\mathrm{n}]{\mathrm{b}}}{2}\right)}^{\mathrm{n}}={\mathrm{e}}^{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\ln {\left(\frac{\sqrt[\mathrm{n}]{\mathrm{a}}+\sqrt[\mathrm{n}]{\mathrm{b}}}{2}\right)}^{\mathrm{n}}}$$$$={\mathrm{e}}^{\underset{\mathrm{x}\to 0^{+}}{\lim }\left(\frac{\ln ({\mathrm{a}}^{\mathrm{x}}+{\mathrm{b}}^{\mathrm{x}})-\ln 2}{\mathrm{x}}\right)};\mathrm{where}\mathrm{n}=\frac{1}{\mathrm{x}}$$$$={\mathrm{e}}^{\underset{x\to 0^{+}}{\lim }\left(\frac{\left(\ln \mathrm{a}\right).{\mathrm{a}}^{\mathrm{x}}+\left(\ln \mathrm{b}\right).{\mathrm{b}}^{\mathrm{x}}}{{\mathrm{a}}^{\mathrm{x}}+{\mathrm{b}}^{\mathrm{x}}}\right)}$$$$={\mathrm{e}}^{\left(\frac{\ln \mathrm{a}+\ln \mathrm{b}}{2}\right)}={\mathrm{e}}^{\ln {\left(\mathrm{ab}\right)}^{\frac{1}{2}}}={\left(\mathrm{ab}\right)}^{\frac{1}{2}}=\sqrt{\mathrm{ab}}$$$$$$

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