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Question Number 129587 by Adel last updated on 16/Jan/21

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

Answered by Dwaipayan Shikari last updated on 16/Jan/21

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

Answered by Ar Brandon last updated on 16/Jan/21

$$\mathcal{L}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{(1-\frac{1}{\mathrm{x}})}^{\mathrm{x}}$$$$\ln \mathcal{L}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim xln}(1-\frac{1}{\mathrm{x}})$$$$=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }-{\mathrm{x}}^2\cdot \frac{1}{{\mathrm{x}}^2}\cdot \frac{\mathrm{x}}{\mathrm{x}-1}=-1$$$$\mathcal{L}={\mathrm{e}}^{-1}=\frac{1}{\mathrm{e}}$$

Answered by physicstutes last updated on 16/Jan/21

$$\underset{x\to \mathrm{\infty }}{\lim }{(1-\frac{1}{x})}^x$$$$\mathrm{let}-\frac{1}{x}=\frac{1}{m}$$$$m=-x\Rightarrow x\to \mathrm{\infty },m\to -\mathrm{\infty }$$$$\underset{x\to -\mathrm{\infty }}{\lim }{(1+\frac{1}{m})}^{-m}=\underset{x\to -\mathrm{\infty }}{\lim }{\left[{(1+\frac{1}{m})}^m\right]}^{-1}=e^{-1}$$

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