mathop {lim }limits_{x to 1} frac{{sqrt {1 - | vedclass

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(D) Let $f(x) = \frac{\sqrt{1 - \cos 2(x - 1)}}{x - 1}$.We know that $1 - \cos 2	heta = 2\sin^2 	heta$,so $1 - \cos 2(x - 1) = 2\sin^2(x - 1)$.Thus,

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Does not exist because left hand limit is not equal to right hand limit

Vedclass · Answer

(D) Let $f(x) = \frac{\sqrt{1 - \cos 2(x - 1)}}{x - 1}$.We know that $1 - \cos 2	heta = 2\sin^2 	heta$,so $1 - \cos 2(x - 1) = 2\sin^2(x - 1)$.Thus,$f(x) = \frac{\sqrt{2\sin^2(x - 1)}}{x - 1} = \frac{\sqrt{2}|\sin(x - 1)|}{x - 1}$.Right Hand Limit $(RHL)$: $\mathop {\lim }\limits_{x 	o 1^+} f(x) = \mathop {\lim }\limits_{h 	o 0^+} \frac{\sqrt{2}|\sin h|}{h} = \mathop {\lim }\limits_{h 	o 0^+} \sqrt{2} \frac{\sin h}{h} = \sqrt{2}$.Left Hand Limit $(LHL)$: $\mathop {\lim }\limits_{x 	o 1^-} f(x) = \mathop {\lim }\limits_{h 	o 0^+} \frac{\sqrt{2}|\sin(-h)|}{-h} = \mathop {\lim }\limits_{h 	o 0^+} \frac{\sqrt{2}\sin h}{-h} = -\sqrt{2}$.Since $	ext{RHL} 
eq 	ext{LHL}$,the limit does not exist.

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

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