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$\lim\limits_{x\to \large\frac{\pi}{4}}\large\frac{\cos x-\sin x}{(\Large\frac{\pi}{4}-\normalsize x)(\large\cos x+\sin x)}$ is

$(a)\;0\qquad(b)\;1\qquad(c)\;-1\qquad(d)\;None\;of\;these$

Can you answer this question?
 
 

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Given limit=$\lim\limits_{x\to \Large\frac{\pi}{4}}\large\frac{1}{(\Large\frac{\pi}{4}-x)}\big(\large\frac{1-\tan x}{1+\tan x}\big)$
$\Rightarrow \lim\limits_{x\to \large\frac{\pi}{4}}\large\frac{\tan(\pi/4-x)}{(\pi/4)-x}$
Substitute $x=\large\frac{\pi}{4}$$+\theta$
So that when $x=\large\frac{\pi}{4}$ ,$\theta\to 0$
$\Rightarrow \lim\limits_{\theta\to 0}\large\frac{\tan(-\theta)}{-\theta}$
$\Rightarrow \lim\limits_{\theta\to 0}\large\frac{\tan(\theta)}{\theta}$
$\Rightarrow 1$
Hence (b) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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