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Question Number 60716 by peter frank last updated on 24/May/19

Commented by maxmathsup by imad last updated on 25/May/19

$$letA\left(x\right)={\left(\frac{x}{x+1}\right)}^x\Rightarrow A\left(x\right)={(1-\frac{x+1}{x})}^x=e^{xln(1-\frac{1}{x+1})}$$$$butwehaveforu\in V\left(0\right)ln(1-u)\sim -u\Rightarrow ln(1-\frac{1}{x+1})\sim -\frac{1}{x+1}forx\in V(+\mathrm{\infty })$$$$\Rightarrow xln(1-\frac{1}{x+1})\sim -\frac{x}{x+1}\to -1(x\to +\mathrm{\infty })\Rightarrow {lim}_{x\to +\mathrm{\infty }}A\left(x\right)=e^{-1}=\frac{1}{e}$$$$★{lim}_{x\to +\mathrm{\infty }}{\left(\frac{x}{x+1}\right)}^x=\frac{1}{e}★.$$

Answered by Smail last updated on 24/May/19

$${\left(\frac{x}{1+x}\right)}^x={\left(\frac{1}{1+1/x}\right)}^x={(1+\frac{1}{x})}^{-x}=e^{-xln(1+1/x)}$$$$Lett=1/x$$$$e^{-xln(1+1/x)}=e^{-\frac{ln(1+t)}{t}}$$$$ln(1+t)\underset{0}{\sim }t$$$$So{e}^{-\frac{ln(1+t)}{t}}\underset{0}{\sim }{e}^{-\frac{t}{t}}=\frac{1}{e}$$$$Thus,{\underset{x\to \mathrm{\infty }}{lime}}^{-xln(1+1/x)}=\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x}{1+x}\right)}^x=\frac{1}{e}$$

Commented by peter frank last updated on 24/May/19

$$thankyou$$

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