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Question Number 211750 by Alijumaaxyz last updated on 19/Sep/24

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{a^{1/x}+b^{1/x}}{2}\right)}^x;(a,b)\in {\mathbb{R}}_{+}^{\ast }$$

Answered by mehdee7396 last updated on 19/Sep/24

$$li\underset{x\to \mathrm{\infty }}{m}b{\left(\frac{{\left(\frac{a}{b}\right)}^{\frac{1}{x}}+1}{2}\right)}^x$$$$★li\underset{x\to \mathrm{\infty }}{m}{\left(\frac{{\left(\frac{a}{b}\right)}^{\frac{1}{x}}+1}{2}\right)}^x$$$$=li\underset{x\to \mathrm{\infty }}{m}(\frac{{\left(\frac{a}{b}\right)}^{\frac{1}{x}}+1}{2}-1)x$$$$=li\underset{x\to \mathrm{\infty }}{m}\left(\frac{{\left(\frac{a}{b}\right)}^{\frac{1}{x}}-1}{2}\right)x$$$$=li\underset{x\to \mathrm{\infty }}{m}\frac{{\left(\frac{a}{b}\right)}^{\frac{1}{x}}-1}{\frac{2}{x}}$$$$=li\underset{x\to \mathrm{\infty }}{m}\frac{-\frac{1}{x^2}{\left(\frac{a}{b}\right)}^{\frac{1}{x}}ln\left(\frac{a}{b}\right)}{-\frac{2}{x^2}}=\frac{1}{2}ln\left(\frac{a}{b}\right)$$$$\Rightarrow li\underset{x\to \mathrm{\infty }}{m}{\left(\frac{{\left(\frac{a}{b}\right)}^{\frac{1}{x}}+1}{2}\right)}^x=e^{\frac{1}{2}ln\left(\frac{a}{b}\right)}=\sqrt{\frac{a}{b}}$$$$\Rightarrow li\underset{x\to \mathrm{\infty }}{m}b{\left(\frac{{\left(\frac{a}{b}\right)}^{\frac{1}{x}}+1}{2}\right)}^x=\sqrt{ab}✓$$$$$$

Commented by Alijumaaxyz last updated on 19/Sep/24

$$✓thankyousir$$

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