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$\lim\limits_{x\to \large\frac{\pi}{4}}\large\frac{\sqrt 2\cos x-1}{\cos x-1}$=

$(a)\;\large\frac{1}{\sqrt 2}$$\qquad(b)\;\large\frac{1}{2}$$\qquad(c)\;\large\frac{1}{2\sqrt 2}\qquad$$(d)\;1$

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$\lim\limits_{x\to \large\frac{\pi}{4}}\large\frac{\sqrt 2\cos x-1}{\cot x-1}\qquad$$(0/0)$ Applying L Hospital's Rule
$\Rightarrow \lim\limits_{x\to\Large\frac{\pi}{4}}\large\frac{-\sqrt 2\sin x}{-cosec^2x}$
$\Rightarrow \sqrt 2\lim\limits_{x\to \large\frac{\pi}{4}}\sin^3x$
$\Rightarrow \sqrt 2\big(\large\frac{1}{\sqrt 2}\big)^3$
$\Rightarrow \large\frac{1}{2}$
Hence (b) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 

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