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Question Number 66019 by Rio Michael last updated on 07/Aug/19

$$\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{2}{x})}^x=$$

Commented by mathmax by abdo last updated on 07/Aug/19

$$letf\left(x\right)={(1+\frac{2}{x})}^x\Rightarrow f\left(x\right)=e^{xln(1+\frac{2}{x})}$$$$wehave{ln}^{{}^{\prime }}(1+u)=\frac{1}{1+u}=1-u+o\left(u^2\right)foru\in V\left(0\right)\Rightarrow$$$$ln(1+u)=u-\frac{u^2}{2}+o\left(u^3\right)\Rightarrow ln(1+\frac{2}{x})=\frac{2}{x}-\frac{4}{2x^2}+o\left(\frac{1}{x^3}\right)(x\to +\mathrm{\infty })$$$$\Rightarrow xln(1+\frac{2}{x})=2-\frac{2}{x}+o\left(\frac{1}{x^2}\right)\Rightarrow f\left(x\right)=e^{2-\frac{2}{x}+o\left(\frac{1}{x^2}\right)}$$$$=e^2.e^{-\frac{2}{x}+o\left(\frac{1}{x^2}\right)}=e^2(1-\frac{2}{x}+o\left(\frac{1}{x^2}\right))\Rightarrow {lim}_{x\to +\mathrm{\infty }}f\left(x\right)=e^2.$$

Answered by Tanmay chaudhury last updated on 07/Aug/19

$$t=\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{2}{x})}^x$$$$y=\frac{1}{x}asx\to \mathrm{\infty }y\to 0$$$$t=\underset{y\to 0}{\lim }{(1+2y)}^{\frac{1}{y}}$$$$lnt=\underset{y\to 0}{\lim }\frac{ln(1+2y)}{2y}\times 2$$$$lnt=1\times 2[formula\underset{x\to 0}{\lim }\frac{ln(1+x)}{x}=1]$$$$t=e^{2}answer$$

Answered by kaivan.ahmadi last updated on 07/Aug/19

$$=e^{{lim}_{x\to \mathrm{\infty }}(1+\frac{2}{x}-1)x}=e^{{lim}_{x\to \mathrm{\infty }}\frac{2}{x}\times x}=e^2$$

Commented by Rio Michael last updated on 07/Aug/19

$$pleaseguyscheckmysolutionasiunderstandit.$$$$\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{2}{x})}^x$$$$let\frac{2}{x}=\frac{1}{m}$$$$\Rightarrow x=2m$$$$suchthatx\to \mathrm{\infty },m\to \mathrm{\infty }$$$$\underset{m\to \mathrm{\infty }}{lim}{\left[{(1+\frac{1}{m})}^m\right]}^2$$$$=e^2$$

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