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Question Number 63964 by meme last updated on 11/Jul/19

$$li{\underset{x\to +\mathrm{\infty }}{me}}^{-xln(1-\frac{1}{x})}$$

Commented by Prithwish sen last updated on 11/Jul/19

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

Commented by turbo msup by abdo last updated on 12/Jul/19

$$chsngement\frac{1}{x}=tgive$$$${lim}_{x\to +\mathrm{\infty }}{e}^{-xln(1-\frac{1}{x})}$$$$={lim}_{t\to 0}{e}^{-\frac{ln(1-t)}{t}}$$$$ln(1-t)\sim -t\Rightarrow \frac{ln(1-t)}{t}\sim -1\Rightarrow$$$${lim}_{t\to 0}{e}^{-\frac{ln(1-t)}{t}}=e$$

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