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$\lim\limits_{x\to 0}\large\frac{\sqrt{1-\cos 2x}}{\sqrt{2x}}$ is

$(a)\;1\qquad(b)\;-1\qquad(c)\;0\qquad(d)\;does\;not\;exist$

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$\lim\limits_{x\to 0}\large\frac{\sqrt{1-\cos 2x}}{\sqrt{2x}}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\sqrt{1-(1-2\sin^2x)}}{\sqrt{2x}}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\sqrt{2\sin^2x}}{\sqrt 2x}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{|\sin x|}{x}$
$\Rightarrow $ The limit of above does not exist as LHS=-1$\neq RHL=1$
Hence (d) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 

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