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

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

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