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Question Number 107882 by Ar Brandon last updated on 13/Aug/20

$$\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{\left(\frac{{\mathrm{a}}^{1/\mathrm{x}}+{\mathrm{b}}^{1/\mathrm{x}}}{2}\right)}^{\mathrm{x}}$$

Commented by bemath last updated on 13/Aug/20

$$coolll$$

Commented by bemath last updated on 13/Aug/20

$$\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)}^x={\left(\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)}^x.\underset{n\to \mathrm{\infty }}{\lim }\left(1\right)$$$$={\left(\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)}^x$$

Commented by Ar Brandon last updated on 13/Aug/20

�� Sorry I meant x instead of n

Commented by bemath last updated on 13/Aug/20

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Commented by kaivan.ahmadi last updated on 13/Aug/20

$${lim}_{x\to \mathrm{\infty }}(\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}-1)\left(x\right)={lim}_{x\to \mathrm{\infty }}\left(\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}-2}{2}\right)\left(x\right)$$$$={lim}_{x\to \mathrm{\infty }}\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}-2}{\frac{2}{x}}\stackrel{Hop}{=}{lim}_{x\to \mathrm{\infty }}\frac{-\frac{1}{x^2}{a}^{\frac{1}{x}}lna-\frac{1}{x^2}{b}^{\frac{1}{x}}lnb}{\frac{-2}{x^2}}=$$$${lim}_{x\to \mathrm{\infty }}\frac{a^{\frac{1}{x}}lna+b^{\frac{1}{x}}lnb}{2}=\frac{lna+lnb}{2}$$$$$$

Answered by pticantor last updated on 14/Aug/20

$$$$$$\sqrt{ab}$$

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