Question Number 107222 by bobhans last updated on 09/Aug/20
Answered by john santu last updated on 09/Aug/20
![⊡JS⊡ lim_(x→∞) (7^(2x) (1+(1/7^x )))^(2/x) = 7^4 ×lim_(x→∞) (1+(1/7^x ))^(2/x) = 7^4 ×[lim_(x→∞) (1+(1/7^x ))^7^x ]^(2/(x.7^x )) = 7^4 × e^(lim_(x→∞) ((2/(x.7^x )))) = 7^4 ×e^0 = 7^4](https://www.tinkutara.com/question/Q107223.png)
$$\:\:\:\:\:\:\:\boxdot\mathrm{JS}\boxdot \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{7}^{\mathrm{2x}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)\right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\:\mathrm{7}^{\mathrm{4}} \:×\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)^{\frac{\mathrm{2}}{\mathrm{x}}} \\ $$$$=\:\mathrm{7}^{\mathrm{4}} \:×\left[\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)^{\mathrm{7}^{\mathrm{x}} } \:\right]^{\frac{\mathrm{2}}{\mathrm{x}.\mathrm{7}^{\mathrm{x}} }} \\ $$$$=\:\mathrm{7}^{\mathrm{4}} \:×\:\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{x}.\mathrm{7}^{\mathrm{x}} }\right)} =\:\mathrm{7}^{\mathrm{4}} \:×\mathrm{e}^{\mathrm{0}} \:=\:\mathrm{7}^{\mathrm{4}} \\ $$
Answered by Dwaipayan Shikari last updated on 09/Aug/20
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{7}^{\mathrm{x}} +\mathrm{7}^{\mathrm{2x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\mathrm{7}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}7}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)^{\frac{\mathrm{7}^{\mathrm{x}} }{\mathrm{7}^{\mathrm{x}} }.\frac{\mathrm{2}}{\mathrm{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{7}^{\mathrm{4}} \mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x7}^{\mathrm{x}} }} =\mathrm{7}^{\mathrm{4}} \mathrm{e}^{\mathrm{0}} =\mathrm{7}^{\mathrm{4}} \\ $$
Answered by Dwaipayan Shikari last updated on 09/Aug/20
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{7}^{\mathrm{x}} +\mathrm{7}^{\mathrm{2x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\mathrm{y} \\ $$$$\frac{\mathrm{2}}{\mathrm{x}}\mathrm{log7}^{\mathrm{2x}} +\frac{\mathrm{2}}{\mathrm{x}}\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)=\mathrm{logy} \\ $$$$\mathrm{log7}^{\mathrm{4}} =\mathrm{logy}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\rightarrow\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{7}^{\mathrm{4}} \\ $$
Answered by abdomathmax last updated on 09/Aug/20
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{7}^{\mathrm{x}} \:+\mathrm{7}^{\mathrm{2x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{7}^{\mathrm{x}} \:+\mathrm{7}^{\mathrm{2x}} \right)} \:=\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{7}^{\mathrm{2x}} \left(\mathrm{1}+\mathrm{7}^{−\mathrm{x}} \right)\right)} \\ $$$$=\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x}}\left\{\mathrm{2x}\:\mathrm{ln}\left(\mathrm{7}\right)+\mathrm{ln}\left(\mathrm{1}+\mathrm{7}^{−\mathrm{x}} \right)\right\}} \:=\:\mathrm{e}^{\mathrm{4ln}\left(\mathrm{7}\right)} \:.\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{7}^{−\mathrm{x}} \right)} \\ $$$$\mathrm{but}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{7}^{−\boldsymbol{\mathrm{x}}} \right)\:\sim\mathrm{7}^{−\mathrm{x}} \:\Rightarrow\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{7}^{−\mathrm{x}} \right)\sim\frac{\mathrm{2}.\mathrm{7}^{−\mathrm{x}} }{\mathrm{x}}\:\rightarrow\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{e}^{\left(…\right)} \rightarrow\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{7}^{\mathrm{4}} \\ $$
Answered by 1549442205PVT last updated on 10/Aug/20
![This is the form ∞^0 lim_(x→∞) (7^x +7^(2x) )^(2/x) =lim_(x→∞) [7^(2x) (1+((7^x /7^(2x) ))^x ]^(2/x) = =7^(2x.(2/x)) .lim_(x→∞) [(1+((1/7))^x ]^(2/x) =7^4 .lim_(x→∞) [(1+((1/7))^x ]^(2/x) = 7^4 (1+0)^0 =7 since ((1/7))^x →0 when x→∞ (2/x)→0 whenx→∞.Answer is 7^4](https://www.tinkutara.com/question/Q107306.png)
$$\mathrm{This}\:\mathrm{is}\:\mathrm{the}\:\mathrm{form}\:\infty^{\mathrm{0}} \\ $$$$\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{7}^{\mathrm{x}} +\mathrm{7}^{\mathrm{2x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{7}^{\mathrm{2x}} \left(\mathrm{1}+\left(\frac{\mathrm{7}^{\mathrm{x}} }{\mathrm{7}^{\mathrm{2x}} }\right)^{\mathrm{x}} \right]^{\frac{\mathrm{2}}{\mathrm{x}}} =\right. \\ $$$$=\mathrm{7}^{\mathrm{2x}.\frac{\mathrm{2}}{\mathrm{x}}} .\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{7}}\right)^{\mathrm{x}} \right]^{\frac{\mathrm{2}}{\mathrm{x}}} =\mathrm{7}^{\mathrm{4}} .\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{7}}\right)^{\mathrm{x}} \right]^{\frac{\mathrm{2}}{\mathrm{x}}} =\right.\right. \\ $$$$\mathrm{7}^{\mathrm{4}} \left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{0}} =\mathrm{7}\:\mathrm{since}\:\left(\frac{\mathrm{1}}{\mathrm{7}}\right)^{\mathrm{x}} \rightarrow\mathrm{0}\:\mathrm{when}\:\mathrm{x}\rightarrow\infty \\ $$$$\frac{\mathrm{2}}{\mathrm{x}}\rightarrow\mathrm{0}\:\mathrm{whenx}\rightarrow\infty.\boldsymbol{\mathrm{Answer}}\:\boldsymbol{\mathrm{is}}\:\mathrm{7}^{\mathrm{4}} \: \\ $$