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Question Number 66084 by F_Nongue last updated on 09/Aug/19

How to solve this limit? lim_(x→∞) (7x+(2/x))^x

$${How}\:{to}\:{solve}\:{this}\:{limit}? \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}} \\ $$

Commented by mathmax by abdo last updated on 09/Aug/19

let A(x) =(7x+(2/x))^x ⇒A(x) =(7x)^x (1+(2/(7x^2 )))^x =(7x)^x e^(xln(1+(2/(7x^2 )))) but ln(1+(2/(7x^2 )))∼(2/(7x^2 )) ⇒xln(1+(2/(7x^2 )))∼(2/(7x)) ⇒ A(x) ∼(7x)^x ×(2/(7x)) =2(7x)^(x−1) =2 e^((x−1)ln(7x)) →+∞ (x→+∞) ⇒ lim_(x→+∞) (7x+(2/x))^x =+∞

$${let}\:{A}\left({x}\right)\:=\left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}} \:\Rightarrow{A}\left({x}\right)\:=\left(\mathrm{7}{x}\right)^{{x}} \left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{7}{x}^{\mathrm{2}} }\right)^{{x}} \\ $$$$=\left(\mathrm{7}{x}\right)^{{x}} \:{e}^{{xln}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{7}{x}^{\mathrm{2}} }\right)} \:\:\:\:\:{but}\:\:{ln}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{7}{x}^{\mathrm{2}} }\right)\sim\frac{\mathrm{2}}{\mathrm{7}{x}^{\mathrm{2}} }\:\Rightarrow{xln}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{7}{x}^{\mathrm{2}} }\right)\sim\frac{\mathrm{2}}{\mathrm{7}{x}}\:\Rightarrow \\ $$$${A}\left({x}\right)\:\sim\left(\mathrm{7}{x}\right)^{{x}} \:×\frac{\mathrm{2}}{\mathrm{7}{x}}\:=\mathrm{2}\left(\mathrm{7}{x}\right)^{{x}−\mathrm{1}} \:=\mathrm{2}\:{e}^{\left({x}−\mathrm{1}\right){ln}\left(\mathrm{7}{x}\right)} \:\rightarrow+\infty\:\:\left({x}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} \left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}} \:=+\infty\: \\ $$

Answered by MJS last updated on 09/Aug/19

(7x+(2/x))^x =(((7x^2 +2)/x))^x =(((7x^2 +2)^x )/x^x ) lim_(x→∞) (((7x^2 +2)^x )/x^x )>lim_(x→∞) (((7x^2 )^x )/x^x )=lim_(x→∞) 7^x x^x =+∞

$$\left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}} =\left(\frac{\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}}{{x}}\right)^{{x}} =\frac{\left(\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}\right)^{{x}} }{{x}^{{x}} } \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}\right)^{{x}} }{{x}^{{x}} }>\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{7}{x}^{\mathrm{2}} \right)^{{x}} }{{x}^{{x}} }=\underset{{x}\rightarrow\infty} {\mathrm{lim}7}^{{x}} {x}^{{x}} =+\infty \\ $$

Commented by Prithwish sen last updated on 09/Aug/19

(2/x)→0 as x→∞ ∴ (7x+(2/x))^x →7^x x^x

$$\:\frac{\mathrm{2}}{\mathrm{x}}\rightarrow\mathrm{0}\:\mathrm{as}\:\mathrm{x}\rightarrow\infty\:\therefore\:\left(\mathrm{7x}+\frac{\mathrm{2}}{\mathrm{x}}\right)^{\mathrm{x}} \rightarrow\mathrm{7}^{\mathrm{x}} \mathrm{x}^{\mathrm{x}} \\ $$

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