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Solution

$1-\cos 2x=2\sin^2 x \Rightarrow \dfrac{\sqrt{1-\cos 2x}}{x}=\sqrt{2},\dfrac{|\sin x|}{x}$. As $x\to0^+$, this $\to\sqrt{2}$; as $x\to0^-$, it $\to -\sqrt{2}$. Thus two-sided limit does not exist; the right-hand limit is $\sqrt{2}$.