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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

$$\:\:\:\:\:\:\:\boxdot\mathrm{JS}\boxdot \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{7}^{\mathrm{2x}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)\right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\:\mathrm{7}^{\mathrm{4}} \:×\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)^{\frac{\mathrm{2}}{\mathrm{x}}} \\ $$$$=\:\mathrm{7}^{\mathrm{4}} \:×\left[\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)^{\mathrm{7}^{\mathrm{x}} } \:\right]^{\frac{\mathrm{2}}{\mathrm{x}.\mathrm{7}^{\mathrm{x}} }} \\ $$$$=\:\mathrm{7}^{\mathrm{4}} \:×\:\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{x}.\mathrm{7}^{\mathrm{x}} }\right)} =\:\mathrm{7}^{\mathrm{4}} \:×\mathrm{e}^{\mathrm{0}} \:=\:\mathrm{7}^{\mathrm{4}} \\ $$

Answered by Dwaipayan Shikari last updated on 09/Aug/20

lim_(x→∞) (7^x +7^(2x) )^(2/x) =7^4 (1+(1/7^x ))^(2/x) =lim_(x→∞) 7^4 (1+(1/7^x ))^((7^x /7^x ).(2/x))                                                                =7^4 e^(2/(x7^x )) =7^4 e^0 =7^4

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{7}^{\mathrm{x}} +\mathrm{7}^{\mathrm{2x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\mathrm{7}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}7}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)^{\frac{\mathrm{7}^{\mathrm{x}} }{\mathrm{7}^{\mathrm{x}} }.\frac{\mathrm{2}}{\mathrm{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{7}^{\mathrm{4}} \mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x7}^{\mathrm{x}} }} =\mathrm{7}^{\mathrm{4}} \mathrm{e}^{\mathrm{0}} =\mathrm{7}^{\mathrm{4}} \\ $$

Answered by Dwaipayan Shikari last updated on 09/Aug/20

lim_(x→∞) (7^x +7^(2x) )^(2/x) =y  (2/x)log7^(2x) +(2/x)log(1+(1/7^x ))=logy  log7^4 =logy                (1/7^x )→0  y=7^4

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{7}^{\mathrm{x}} +\mathrm{7}^{\mathrm{2x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\mathrm{y} \\ $$$$\frac{\mathrm{2}}{\mathrm{x}}\mathrm{log7}^{\mathrm{2x}} +\frac{\mathrm{2}}{\mathrm{x}}\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)=\mathrm{logy} \\ $$$$\mathrm{log7}^{\mathrm{4}} =\mathrm{logy}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\rightarrow\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{7}^{\mathrm{4}} \\ $$

Answered by abdomathmax last updated on 09/Aug/20

let f(x) =(7^x  +7^(2x) )^(2/x)   ⇒f(x) =e^((2/x)ln(7^x  +7^(2x) ))  =e^((2/x)ln(7^(2x) (1+7^(−x) )))   =e^((2/x){2x ln(7)+ln(1+7^(−x) )})  = e^(4ln(7))  .e^((2/x)ln(1+7^(−x) ))   but ln(1+7^(−x) ) ∼7^(−x)  ⇒(2/x)ln(1+7^(−x) )∼((2.7^(−x) )/x) →0 ⇒  e^((...)) →1 ⇒lim_(x→+∞) f(x) =7^4

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{7}^{\mathrm{x}} \:+\mathrm{7}^{\mathrm{2x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{7}^{\mathrm{x}} \:+\mathrm{7}^{\mathrm{2x}} \right)} \:=\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{7}^{\mathrm{2x}} \left(\mathrm{1}+\mathrm{7}^{−\mathrm{x}} \right)\right)} \\ $$$$=\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x}}\left\{\mathrm{2x}\:\mathrm{ln}\left(\mathrm{7}\right)+\mathrm{ln}\left(\mathrm{1}+\mathrm{7}^{−\mathrm{x}} \right)\right\}} \:=\:\mathrm{e}^{\mathrm{4ln}\left(\mathrm{7}\right)} \:.\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{7}^{−\mathrm{x}} \right)} \\ $$$$\mathrm{but}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{7}^{−\boldsymbol{\mathrm{x}}} \right)\:\sim\mathrm{7}^{−\mathrm{x}} \:\Rightarrow\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{7}^{−\mathrm{x}} \right)\sim\frac{\mathrm{2}.\mathrm{7}^{−\mathrm{x}} }{\mathrm{x}}\:\rightarrow\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{e}^{\left(...\right)} \rightarrow\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{7}^{\mathrm{4}} \\ $$

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

$$\mathrm{This}\:\mathrm{is}\:\mathrm{the}\:\mathrm{form}\:\infty^{\mathrm{0}} \\ $$$$\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{7}^{\mathrm{x}} +\mathrm{7}^{\mathrm{2x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{7}^{\mathrm{2x}} \left(\mathrm{1}+\left(\frac{\mathrm{7}^{\mathrm{x}} }{\mathrm{7}^{\mathrm{2x}} }\right)^{\mathrm{x}} \right]^{\frac{\mathrm{2}}{\mathrm{x}}} =\right. \\ $$$$=\mathrm{7}^{\mathrm{2x}.\frac{\mathrm{2}}{\mathrm{x}}} .\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{7}}\right)^{\mathrm{x}} \right]^{\frac{\mathrm{2}}{\mathrm{x}}} =\mathrm{7}^{\mathrm{4}} .\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{7}}\right)^{\mathrm{x}} \right]^{\frac{\mathrm{2}}{\mathrm{x}}} =\right.\right. \\ $$$$\mathrm{7}^{\mathrm{4}} \left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{0}} =\mathrm{7}\:\mathrm{since}\:\left(\frac{\mathrm{1}}{\mathrm{7}}\right)^{\mathrm{x}} \rightarrow\mathrm{0}\:\mathrm{when}\:\mathrm{x}\rightarrow\infty \\ $$$$\frac{\mathrm{2}}{\mathrm{x}}\rightarrow\mathrm{0}\:\mathrm{whenx}\rightarrow\infty.\boldsymbol{\mathrm{Answer}}\:\boldsymbol{\mathrm{is}}\:\mathrm{7}^{\mathrm{4}} \: \\ $$

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