mathop {lim }limits_{x to 0} frac{{sqrt {frac | vedclass

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Question

(D) We know that $1 - \cos 2x = 2 \sin^2 x$.Substituting this into the expression,we get $\sqrt{\frac{1}{2}(2 \sin^2 x)} = \sqrt{\sin^2 x} = |\sin x|$

Vedclass · Accepted Answer

$	ext{Does not exist}$

Vedclass · Answer

(D) We know that $1 - \cos 2x = 2 \sin^2 x$.Substituting this into the expression,we get $\sqrt{\frac{1}{2}(2 \sin^2 x)} = \sqrt{\sin^2 x} = |\sin x|$.Thus,the limit becomes $\mathop {\lim }\limits_{x 	o 0} \frac{|\sin x|}{x}$.For the right-hand limit $(x 	o 0^+)$,$|\sin x| = \sin x$,so $\mathop {\lim }\limits_{x 	o 0^+} \frac{\sin x}{x} = 1$.For the left-hand limit $(x 	o 0^-)$,$|\sin x| = -\sin x$,so $\mathop {\lim }\limits_{x 	o 0^-} \frac{-\sin x}{x} = -1$.Since the left-hand limit and right-hand limit are not equal,the limit does not exist.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $

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