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Question Number 104746 by bobhans last updated on 23/Jul/20
lim_(x→∞) (((1/2))^(3x) +((1/2))^x )^(1/x^2 )
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}{x}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \: \\ $$
Answered by Dwaipayan Shikari last updated on 23/Jul/20
lim_(x→∞) (((1/2))^(3x) +((1/2))^x )^(1/x^2 ) =y (1/x^2 )log(((1/2))^(3x) +((1/2))^x )=logy (1/x^2 )log(1+z−1).((z−1)/(z−1))=logy {take z→0 as ((1/2))^x →0 ((((1/2))^(3x) +((1/2))^x −1)/x^2 )=logy logy=0 y=1
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}{x}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} ={y} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{log}\left(\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}{x}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right)={logy} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{log}\left(\mathrm{1}+{z}−\mathrm{1}\right).\frac{{z}−\mathrm{1}}{{z}−\mathrm{1}}={logy}\:\:\:\:\:\:\:\:\:\left\{{take}\:{z}\rightarrow\mathrm{0}\:\:\:\:{as}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \rightarrow\mathrm{0}\right. \\ $$$$\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}{x}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} −\mathrm{1}}{{x}^{\mathrm{2}} }={logy} \\ $$$${logy}=\mathrm{0} \\ $$$${y}=\mathrm{1} \\ $$
Answered by mathmax by abdo last updated on 23/Jul/20
f(x) ={((1/2))^(3x) +((1/2))^x }^(1/x^2 ) ⇒ln(f(x) =(1/x^2 )ln{((1/2))^(3x) +((1/2))^x } =(1/x^2 )ln{((1/2))^x (((1/2))^(2x) +1)} =−(1/x)ln(2) +(1/x^2 )ln(1+(1/4^x )) ⇒ ln(f(x)) ∼−((ln(2))/x) +(1/(x^2 .4^x )) (x→+∞) ⇒lim_(x→+∞) ln(f(x)) =0 ⇒lim_(x→+∞) f(x)=1
$$\mathrm{f}\left(\mathrm{x}\right)\:=\left\{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3x}} \:+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{x}} \right\}^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \:\:\Rightarrow\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left\{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3x}} \:+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{x}} \right\}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left\{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{x}} \left(\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2x}} \:+\mathrm{1}\right)\right\}\:=−\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{x}} }\right)\:\Rightarrow \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\:\sim−\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{x}}\:+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} .\mathrm{4}^{\mathrm{x}} }\:\left(\mathrm{x}\rightarrow+\infty\right)\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\:=\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{1} \\ $$

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