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Question Number 74782 by naka3546 last updated on 30/Nov/19

Commented by abdomathmax last updated on 01/Dec/19

$$letf\left(x\right)={\{{\left(\frac{1}{8}\right)}^x+{\left(\frac{1}{2}\right)}^x\}}^{\frac{1}{x^2}}\Rightarrow$$$$ln\left(f\left(x\right)\right)=\frac{1}{x^2}ln\{{\left(\frac{1}{8}\right)}^x+{\left(\frac{1}{2}\right)}^x\}$$$$=\frac{1}{x}ln\left(\frac{1}{8}\right)+\frac{1}{x^2}ln\{1+4^x\}$$$$=\frac{-3ln\left(2\right)}{x}+\frac{2ln\left(2\right)}{x}+ln\{1+4^{-x}\}$$$$\Rightarrow f\left(x\right)\sim -\frac{ln\left(2\right)}{x}+4^{-x}\to 0(x\to +\mathrm{\infty })\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}f\left(x\right)=1$$

Commented by abdomathmax last updated on 24/Dec/19

$$ln\left(f\left(x\right)\right)\sim -\frac{ln\left(2\right)}{x}+4^{-x}(x\to +\mathrm{\infty })$$

Answered by MJS last updated on 30/Nov/19

$${(\frac{1}{2^{3x}}+\frac{1}{2^x})}^{\frac{1}{x^2}}={\left(\frac{1}{8^x}(1+4^x)\right)}^{\frac{1}{x^2}}=\frac{\sqrt[x^2]{1+4^x}}{\sqrt[x]{8}}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\sqrt[x]{8}}=1$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt[x^2]{1+4^x}\sim \underset{x\to \mathrm{\infty }}{\lim }\sqrt[x]{4}=1$$$$\Rightarrow \mathrm{answer}\mathrm{is}\left(\mathrm{C}\right)$$

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