$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} x}$ is
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$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} x}$ is

(A) 1

(B) -1

(C) 0

(D) Does not exist

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Best answer

SOLUTION —

$\begin{array}{l}\text {} \lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} x}=\lim _{x \rightarrow 0} \frac{\sqrt{2 \sin ^2 x}}{\sqrt{2} x} \\=\lim _{x \rightarrow 0} \frac{|\sin x|}{x}=f(x) \\\mathrm{LHL}=\lim _{h \rightarrow 0} \frac{|\sin (0-h)|}{-h}=-1 \\\mathrm{RHL}=\lim _{h \rightarrow 0} \frac{|\sin (0+h)|}{h}=1 \\\therefore \quad \mathrm{LHL} \neq \mathrm{RHL} \\\end{array}$

So, The correct option will be (D).

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