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$\lim\limits_{x\to 1}(1-x)\tan\large\frac{\pi x}{2}$=__________

$\begin{array}{1 1}(a)\;\large\frac{\pi}{2}&(b)\;\pi\\(c)\;\large\frac{2}{\pi}&(d)\;0\end{array}$

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Apply 'L' Hospital rule.We get
$\lim\limits_{x\to 0}\large\frac{f(x)}{g(x)}=$$\lim\limits_{x\to a}\large\frac{f'(x)}{g'(x)}$
If $\large\frac{f(a)}{g(a)}$ is of the form $\large\frac{0}{0}$ or $\large\frac{\infty}{\infty}$ or $0\times \infty$
$\lim\limits_{x\to 1}(1-x)\tan\large\frac{\pi x}{2}$ [from $0\times \infty]$
$\lim\limits_{x\to 1}\large\frac{(1-x)}{\cot(\large\frac{\pi x}{2})}$ [from $0/ 0]$
$\Rightarrow \lim\limits_{x\to 1}\large\frac{-1}{(-\pi/2)cosec^2(\pi x/2)}$
$\Rightarrow \large\frac{2}{\pi}$
Hence (c) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 

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