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Question Number 104746 by bobhans last updated on 23/Jul/20

$$\underset{x\to \mathrm{\infty }}{\lim }{({\left(\frac{1}{2}\right)}^{3x}+{\left(\frac{1}{2}\right)}^x)}^{\frac{1}{x^2}}$$

Answered by Dwaipayan Shikari last updated on 23/Jul/20

$$\underset{x\to \mathrm{\infty }}{\lim }{({\left(\frac{1}{2}\right)}^{3x}+{\left(\frac{1}{2}\right)}^x)}^{\frac{1}{x^2}}=y$$$$\frac{1}{x^2}log({\left(\frac{1}{2}\right)}^{3x}+{\left(\frac{1}{2}\right)}^x)=logy$$$$\frac{1}{x^2}log(1+z-1).\frac{z-1}{z-1}=logy\{takez\to 0as{\left(\frac{1}{2}\right)}^x\to 0$$$$\frac{{\left(\frac{1}{2}\right)}^{3x}+{\left(\frac{1}{2}\right)}^x-1}{x^2}=logy$$$$logy=0$$$$y=1$$

Answered by mathmax by abdo last updated on 23/Jul/20

$$\mathrm{f}\left(\mathrm{x}\right)={\{{\left(\frac{1}{2}\right)}^{3\mathrm{x}}+{\left(\frac{1}{2}\right)}^{\mathrm{x}}\}}^{\frac{1}{{\mathrm{x}}^2}}\Rightarrow \ln (\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^2}\ln \{{\left(\frac{1}{2}\right)}^{3\mathrm{x}}+{\left(\frac{1}{2}\right)}^{\mathrm{x}}\}$$$$=\frac{1}{{\mathrm{x}}^2}\ln \left\{{\left(\frac{1}{2}\right)}^{\mathrm{x}}({\left(\frac{1}{2}\right)}^{2\mathrm{x}}+1)\right\}=-\frac{1}{\mathrm{x}}\ln \left(2\right)+\frac{1}{{\mathrm{x}}^2}\ln (1+\frac{1}{4^{\mathrm{x}}})\Rightarrow$$$$\ln \left(\mathrm{f}\left(\mathrm{x}\right)\right)\sim -\frac{\ln \left(2\right)}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2.4^{\mathrm{x}}}(\mathrm{x}\to +\mathrm{\infty })\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\ln \left(\mathrm{f}\left(\mathrm{x}\right)\right)=0\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=1$$

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