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Question Number 36563 by bshahid010@gmail.com last updated on 03/Jun/18

Commented by prof Abdo imad last updated on 03/Jun/18

$$letusethechangement\frac{a}{x}=t$$$${lim}_{x\to a}{(2-\frac{a}{x})}^{tan\left(\frac{\pi x}{2a}\right)}={lim}_{t\to 1}{(2-t)}^{tan\left(\frac{\pi }{2t}\right)}$$$$={lim}_{t\to 1}{e}^{tan\left(\frac{\pi }{2t}\right)ln(2-t)}forthatletfind$$$${lim}_{t\to 1}tan\left(\frac{\pi }{2t}\right)ln(2-t)changement$$$$1-t=ugive$$$$tan\left(\frac{\pi }{2t}\right)ln(2-t)=tan\left(\frac{\pi }{2(1-u)}\right)ln(1+u)but$$$$ln(1+u)\sim uand\frac{\pi }{2(1-u)}\sim \frac{\pi }{2}(1+u)\Rightarrow$$$$tan(\frac{\pi }{2}+\frac{\pi u}{2})=-\frac{1}{tan\left(\frac{\pi u}{2}\right)}\Rightarrow$$$$tan\left(\frac{\pi }{2(1-u)}\right)ln(1+u)\sim \frac{-u}{tan\left(\frac{\pi }{2}u\right)}$$$$=-\frac{\frac{\pi }{2}u}{\frac{\pi }{2}tan\left(\frac{\pi }{2}u\right)}\sim \frac{-2}{\pi }(u\to o)\Rightarrow$$$${lim}_{x\to a}{(2-\frac{a}{x})}^{tan\left(\frac{\pi x}{2a}\right)}=e^{-\frac{2}{\pi }}.$$

Answered by ajfour last updated on 03/Jun/18

$$L=\underset{x\to a}{\lim }{\left\{{[1+(1-\frac{a}{x})]}^{\frac{1}{1-\frac{a}{x}}}\right\}}^{\frac{1-\frac{a}{x}}{\tan \left(\frac{\pi }{2}(1-\frac{x}{a})\right)}}$$$$\Rightarrow L=e^{\frac{2}{\pi }}.$$

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