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Question Number 66318 by mathmax by abdo last updated on 12/Aug/19

$$find{lim}_{x\to 0}{\left(\frac{1+x}{1-x}\right)}^{\frac{1}{sinx}}$$

Commented by mathmax by abdo last updated on 24/Aug/19

$$letf\left(x\right)={\left(\frac{1+x}{1-x}\right)}^{\frac{1}{sinx}}\Rightarrow f\left(x\right)=e^{\frac{1}{sinx}ln\left(\frac{1+x}{1-x}\right)}but$$$$ln\left(\frac{1+x}{1-x}\right)=ln(1+x)-ln(1-x)\sim x-(-x)=2xandsinx\sim x\Rightarrow$$$$\frac{1}{sinx}ln\left(\frac{1+x}{1-x}\right)\sim \frac{1}{x}\left(2x\right)=2\Rightarrow {lim}_{x\to 0}f\left(x\right)=e^2$$

Commented by kaivan.ahmadi last updated on 12/Aug/19

$$=e^{{lim}_{x\to 0}(\frac{1+x}{1-x}-1)\times \frac{1}{x}}=e^{{lim}_{x\to 0}\left(\frac{2x}{1-x}\right)\times \frac{1}{x}}=e^2$$$$$$

Commented by Mikael last updated on 12/Aug/19

$$\underset{x\to 0}{lim}{[1+(\frac{1+x}{1-x}-1)]}^{\frac{1}{sinx}}=\underset{x\to 0}{lim}{\left[{(1+\frac{2x}{1-x})}^{\frac{1-x}{2x}}\right]}^{\frac{2x}{1-x}.\frac{1}{sinx}}$$$$\underset{x\to 0}{lim}\left[{(1+\frac{2x}{1-x})}^{\frac{1-x}{2x}}\right]=e,then\underset{x\to 0}{lim}\frac{2x}{1-x}.\frac{1}{sinx}=\underset{x\to 0}{lim}\frac{x}{sinx}.\frac{2}{1-x}=1\times 2=2$$$$=e^2$$

Commented by mathmax by abdo last updated on 12/Aug/19

$$thankssir.$$

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