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Question Number 3048 by Syaka last updated on 03/Dec/15

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

Answered by 123456 last updated on 04/Dec/15

$$\mathrm{L}=\underset{x\to -\mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^x$$$$u=\frac{1}{x}$$$$x\to -\mathrm{\infty }\equiv u\to 0^{-}$$$$\mathrm{L}=\underset{u\to 0^{-}}{\lim }{(1+u)}^{\frac{1}{u}}$$$$\mathrm{remember}\mathrm{that}$$$$e=\underset{x\to 0}{\lim }{(1+x)}^{\frac{1}{x}}$$$$\mathrm{so}$$$$\mathrm{L}=e$$

Commented by prakash jain last updated on 04/Dec/15

$$y={(1+x)}^{1/x}$$$$\ln y=\frac{1}{x}\ln (1+x)$$$$\underset{x\to 0}{\lim ln}y=\underset{x\to 0}{\lim }\frac{\ln (1+x)}{x}=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}\ln (1+x)}{\frac{d}{dx}x}=\underset{x\to 0}{\lim }\frac{1}{1+x}=1$$$$\underset{x\to 0}{\lim ln}y=1$$$$\underset{x\to 0}{\lim }y=e^1=e$$

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