The limit lim _{x rightarrow 1} frac{sqrt{1-c | vedclass

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(C) Let $L = \lim _{x ightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$.Using the identity $1-\cos 2	heta = 2\sin^2	heta$,we have:$L = \lim _{x igh

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does not exist

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(C) Let $L = \lim _{x ightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$.Using the identity $1-\cos 2	heta = 2\sin^2	heta$,we have:$L = \lim _{x ightarrow 1} \frac{\sqrt{2\sin^2(x-1)}}{x-1} = \sqrt{2} \lim _{x ightarrow 1} \frac{|\sin(x-1)|}{x-1}$.Let $z = x-1$. As $x ightarrow 1$,$z ightarrow 0$.$L = \sqrt{2} \lim _{z ightarrow 0} \frac{|\sin z|}{z}$.Now,evaluate the one-sided limits:Right-hand limit $(RHL)$: $\sqrt{2} \lim _{z ightarrow 0^+} \frac{\sin z}{z} = \sqrt{2}(1) = \sqrt{2}$.Left-hand limit $(LHL)$: $\sqrt{2} \lim _{z ightarrow 0^-} \frac{-\sin z}{z} = \sqrt{2}(-1) = -\sqrt{2}$.Since $RHL 
eq LHL$,the limit does not exist.

The limit $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$

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