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Question Number 58438 by salahahmed last updated on 23/Apr/19

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{x-1}{x}\right)}^x$$

Commented by mr W last updated on 23/Apr/19

$$=\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{1}{\frac{x}{x-1}}\right)}^x$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\left(\frac{x}{x-1}\right){\left(\frac{x}{x-1}\right)}^{x-1}}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\left(\frac{1}{1-\frac{1}{x}}\right){(1+\frac{1}{x-1})}^{x-1}}$$$$=\frac{1}{1\times e}$$$$=\frac{1}{e}$$

Commented by salahahmed last updated on 23/Apr/19

$$\mathrm{Why}\mathrm{does}{\left(\frac{\frac{1}{x}}{x-1}\right)}^x=\frac{\left(\frac{x}{x-1}\right)}{{\left(\frac{x}{x-1}\right)}^{x-1}}\mathrm{and}{\left(\frac{x}{x-1}\right)}^{x-1}={(1+\frac{1}{x-1})}^{x-1}$$

Commented by mr W last updated on 23/Apr/19

$$itis{\left(\frac{\frac{1}{x}}{x-1}\right)}^x=\frac{\left(\frac{x-1}{x}\right)}{{\left(\frac{x}{x-1}\right)}^{x-1}}=\frac{(1-\frac{1}{x})}{{\left(\frac{x}{x-1}\right)}^{x-1}}$$$${\left(\frac{x}{x-1}\right)}^{x-1}={\left(\frac{x-1+1}{x-1}\right)}^{x-1}={(1+\frac{1}{x-1})}^{x-1}$$

Commented by maxmathsup by imad last updated on 23/Apr/19

$$letA\left(x\right)={\left(\frac{x-1}{x}\right)}^x\Rightarrow A\left(x\right)={(1-\frac{1}{x})}^x\Rightarrow forx>0$$$$ln\left(A\left(x\right)\right)=xln(1-\frac{1}{x})butwehave{ln}^{{}^{\prime }}(1-u)=-\frac{1}{1-u}=-(1+u+o\left(u^2\right))\Rightarrow$$$$ln(1-u)=-u-\frac{u^2}{2}+o\left(u^3\right)\Rightarrow ln(1-\frac{1}{x})=-\frac{1}{x}-\frac{1}{2x^2}+o\left(\frac{1}{x^3}\right)(x\to +\mathrm{\infty })\Rightarrow$$$$xln(1-\frac{1}{x})=-1-\frac{1}{2x}+o\left(\frac{1}{x^2}\right)\Rightarrow {lim}_{x\to +\mathrm{\infty }}xln(1-\frac{1}{x})=-1\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}ln\left(A\left(x\right)\right)=-1\Rightarrow {lim}_{x\to +\mathrm{\infty }}A\left(x\right)=e^{-1}=\frac{1}{e}$$$$$$

Answered by tanmay last updated on 23/Apr/19

$$anotherway$$$$t=\frac{1}{x}$$$$\underset{t\to 0}{\lim }{(1-t)}^{\frac{1}{t}}$$$$y=\underset{t\to 0}{\lim }{(1-t)}^{\frac{1}{t}}$$$$lny=\underset{t\to 0}{\lim }\frac{ln(1-t)}{-t}\times -1$$$$lny=-1$$$$y=e^{-1}=\frac{1}{e}$$

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