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Question Number 121042 by bramlexs22 last updated on 05/Nov/20

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

Answered by bobhans last updated on 05/Nov/20

$$Letw=\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^x$$$$then\ln \left(w\right)=\ln \left(\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^x\right)$$$$\ln \left(w\right)=\underset{x\to \mathrm{\infty }}{\lim }\left(\ln {(1+\frac{1}{x})}^x\right)$$$$\ln \left(w\right)=\underset{x\to \mathrm{\infty }}{\lim }(x.\ln (1+\frac{1}{x}))$$$$\ln \left(w\right)=\underset{x\to \mathrm{\infty }}{\lim }(x.[\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+...])$$$$\ln \left(w\right)=\underset{x\to \mathrm{\infty }}{\lim }(1-\frac{1}{2x}+\frac{1}{3x^2}-\frac{1}{4x^3}+...)$$$$\ln \left(w\right)=1\Rightarrow w=e^1=e.$$

Answered by Dwaipayan Shikari last updated on 05/Nov/20

$$\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^x$$$$=1+\frac{x}{x}+\frac{x(x-1)}{2!x^2}+\frac{x(x-1)(x-2)}{3!x^3}+...$$$$=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}=e$$

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