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Question Number 79236 by jagoll last updated on 23/Jan/20

lim_(x→+∞) x{e−(1+(1/x))^x }=?

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\mathrm{x}\left\{\mathrm{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}} \right\}=? \\ $$

Commented by mathmax by abdo last updated on 23/Jan/20

let A(x)=x{e−(1+(1/x))^x } we have (1+(1/x))^x =e^(xln(1+(1/x))) we have ln^′ (1+u) =(1/(1+u)) =1−u +o(u^2 ) ⇒ ln(1+u) =u−(u^2 /2) +o(u^3 )(u∼o) ⇒ln(1+(1/x))=(1/x)−(1/(2x^2 )) +o((1/x^3 ))(x∼+∞) ⇒ xln(1+(1/x))=1−(1/(2x)) +o((1/x^2 )) ⇒e^(xln(1+(1/x))) =e^(1−(1/(2x))+o((1/x^2 ))) =e(1−(1/(2x)) +o((1/x^2 ))) ⇒e−(1+(1/x))^x =(e/(2x)) +o((1/x^2 )) ⇒ x{e−(1+(1/x))^x } =(e/2) +o((1/x)) ⇒lim_(x→+∞) A(x)=(e/2)

$${let}\:{A}\left({x}\right)={x}\left\{{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \right\}\:\:{we}\:{have}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} ={e}^{{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \\ $$$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{o}\left({u}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)\:={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({u}^{\mathrm{3}} \right)\left({u}\sim{o}\right)\:\Rightarrow{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left({x}\sim+\infty\right)\:\Rightarrow \\ $$$${xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:\Rightarrow{e}^{{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} ={e}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} \\ $$$$={e}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right)\:\Rightarrow{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} =\frac{{e}}{\mathrm{2}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${x}\left\{{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \right\}\:=\frac{{e}}{\mathrm{2}}\:+{o}\left(\frac{\mathrm{1}}{{x}}\right)\:\Rightarrow{lim}_{{x}\rightarrow+\infty} \:{A}\left({x}\right)=\frac{{e}}{\mathrm{2}} \\ $$

Commented by jagoll last updated on 23/Jan/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mathmax by abdo last updated on 24/Jan/20

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

Answered by Smail last updated on 23/Jan/20

x=1/t lim_(t→0) ((e−(1+t)^(1/t) )/t)=lim_(t→0) ((e−e^((ln(1+t))/t) )/t) lim_(t→0) ((e−e^((t−(t^2 /2))(1/t)) )/t)=lim_(t→0) ((e−e(1−t/2))/t)=(e/2)

$${x}=\mathrm{1}/{t} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{e}−\left(\mathrm{1}+{t}\right)^{\mathrm{1}/{t}} }{{t}}=\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{e}−{e}^{\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}} }{{t}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{e}−{e}^{\left({t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right)\frac{\mathrm{1}}{{t}}} }{{t}}=\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{e}−{e}\left(\mathrm{1}−{t}/\mathrm{2}\right)}{{t}}=\frac{{e}}{\mathrm{2}} \\ $$

Commented by jagoll last updated on 23/Jan/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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