Question Number 79236 by jagoll last updated on 23/Jan/20
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\mathrm{x}\left\{\mathrm{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}} \right\}=? \\ $$
Commented by mathmax by abdo last updated on 23/Jan/20
$${let}\:{A}\left({x}\right)={x}\left\{{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \right\}\:\:{we}\:{have}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} ={e}^{{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \\ $$$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{o}\left({u}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)\:={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({u}^{\mathrm{3}} \right)\left({u}\sim{o}\right)\:\Rightarrow{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left({x}\sim+\infty\right)\:\Rightarrow \\ $$$${xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:\Rightarrow{e}^{{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} ={e}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} \\ $$$$={e}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right)\:\Rightarrow{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} =\frac{{e}}{\mathrm{2}{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${x}\left\{{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \right\}\:=\frac{{e}}{\mathrm{2}}\:+{o}\left(\frac{\mathrm{1}}{{x}}\right)\:\Rightarrow{lim}_{{x}\rightarrow+\infty} \:{A}\left({x}\right)=\frac{{e}}{\mathrm{2}} \\ $$
Commented by jagoll last updated on 23/Jan/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mathmax by abdo last updated on 24/Jan/20
$${you}\:{are}\:{welcome}. \\ $$
Answered by Smail last updated on 23/Jan/20
$${x}=\mathrm{1}/{t} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{e}−\left(\mathrm{1}+{t}\right)^{\mathrm{1}/{t}} }{{t}}=\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{e}−{e}^{\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}} }{{t}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{e}−{e}^{\left({t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right)\frac{\mathrm{1}}{{t}}} }{{t}}=\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{e}−{e}\left(\mathrm{1}−{t}/\mathrm{2}\right)}{{t}}=\frac{{e}}{\mathrm{2}} \\ $$
Commented by jagoll last updated on 23/Jan/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$