mathop {lim }limits_{x to 2} left( {frac{{sqr | vedclass

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Question

(D) Let $f(x) = \frac{\sqrt{1 - \cos\{2(x - 2)\}}}{x - 2}$.Using the identity $1 - \cos(2	heta) = 2\sin^2(	heta)$,we get $1 - \cos\{2(x - 2)\} = 2\s

Vedclass · Accepted Answer

$	ext{does not exist}$

Vedclass · Answer

(D) Let $f(x) = \frac{\sqrt{1 - \cos\{2(x - 2)\}}}{x - 2}$.Using the identity $1 - \cos(2	heta) = 2\sin^2(	heta)$,we get $1 - \cos\{2(x - 2)\} = 2\sin^2(x - 2)$.Thus,$f(x) = \frac{\sqrt{2\sin^2(x - 2)}}{x - 2} = \frac{\sqrt{2}|\sin(x - 2)|}{x - 2}$.For the Left Hand Limit $(LHL)$: $\mathop {\lim }\limits_{x 	o 2^-} \frac{\sqrt{2}|\sin(x - 2)|}{x - 2}$. As $x 	o 2^-$,$(x - 2) $LHL = \mathop {\lim }\limits_{x 	o 2^-} \frac{-\sqrt{2}\sin(x - 2)}{x - 2} = -\sqrt{2}(1) = -\sqrt{2}$.For the Right Hand Limit $(RHL)$: $\mathop {\lim }\limits_{x 	o 2^+} \frac{\sqrt{2}|\sin(x - 2)|}{x - 2}$. As $x 	o 2^+$,$(x - 2) > 0$,so $|\sin(x - 2)| = \sin(x - 2)$.$RHL = \mathop {\lim }\limits_{x 	o 2^+} \frac{\sqrt{2}\sin(x - 2)}{x - 2} = \sqrt{2}(1) = \sqrt{2}$.Since $LHL 
eq RHL$,the limit does not exist.

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

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