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I am unable to evaluate the following limit

$${\lim_{x\to 1} (2 - x)^{tan{\frac{\pi x}{2}}}} $$

I have tried to express in the form ${a^x = e^{ln(a)^x}}$ and then expand ${ln}$ using Maclaurin series. But still the answer is not coming. Please give me hints on how to evaluate it.

Thank you! :))

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Is that supposed to be $$\lim_{x\to1}(2-x)^{\tan(\pi x/2)}$$ – Gerry Myerson Feb 28 '14 at 22:17
Take logs, express in form $0/0$, use l'Hopital? – Gerry Myerson Feb 28 '14 at 22:19
I did everything except using the simple l'Hopital rule. I am really stupid! -_- Thanks a lot!! :)) – Abir Mukherjee Feb 28 '14 at 22:23
Now you can post an answer to the problem (the software may ask you to wait a while), and then later you can accept the answer. That helps to clear up the Unanswered Questions list. – Gerry Myerson Feb 28 '14 at 22:25

Setting $y=x-1$, we have \begin{eqnarray} \lim_{x\to1}(2-x)^{\tan\frac{\pi x}{2}}&=&\lim_{y\to0}(1-y)^{\tan\frac{\pi(y+1)}{2}}=\lim_{y\to0}(1-y)^{-\cot\frac{\pi y}{2}}=\lim_{y\to0}\exp\left(-\cot(\frac{\pi y}{2})\ln(1-y)\right)\\ &=&\lim_{y\to0}\exp\left[-\cos(\frac{\pi y}{2})\cdot\frac{y}{\sin(\frac{\pi y}{2})}\cdot\frac{\ln(1-y)}{y}\right]=\exp\left[-1\cdot\frac{1}{\frac\pi2}\cdot(-1)\right]=e^{\frac2\pi}. \end{eqnarray}

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Say $x=1+t$, and $t\to 0$. Then

$$ \tan \big(\tfrac{\pi x}{2}\big)=\tan \big(\tfrac{\pi (1+t)}{2}\big) =\tan \big(\tfrac{\pi}{2}+\tfrac{\pi t}{2}\big)=-\cot\big(\tfrac{\pi t}{2}\big)=-\frac{\cos(\tfrac{\pi t}{2}\big)}{\sin(\tfrac{\pi t}{2}\big)}= -\frac{2}{\pi t}+{\mathcal O}(1). $$ Hence $$ \lim_{x\to 1} (2-x)^{\tan \big(\tfrac{\pi x}{2}\big)}=\lim_{t\to 0}(1-t)^{-\frac{2}{\pi t}}= \lim_{t\to 0}\left((1-t)^{-1/t}\right)^{2/\pi}=\mathrm{e}^{2/\pi}. $$

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Setting $\displaystyle1-x=y,\tan\frac{\pi x}2=\tan\frac\pi2(1-y)=\tan\left(\frac\pi2-\frac{\pi y}2\right)=\cot\frac{\pi y}2$

$${\lim_{x\to 1} (2 - x)^{\tan{\dfrac{\pi x}2}}} =\left(\lim_{y\to0}(1+y)^\dfrac1y\right)^{\lim_{y\to0}\cos\dfrac{\pi y}2\cdot \lim_{y\to0}\dfrac{\dfrac{\pi y}2}{\sin\dfrac{\pi y}2}\cdot\dfrac2\pi}$$


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