A) Equals \[\frac{1}{\sqrt{2}}\]
B) Does not exist
C) Equals \[\sqrt{2}\]
D) Equals \[-\sqrt{2}\]
Correct Answer: B
Solution :
\[\underset{x\to 2}{\mathop{\lim }}\,\left( \frac{\sqrt{1-\cos 2(x-2)}}{x-2} \right)\] \[\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{2}\left| \sin (x-2) \right|}{x-2}\]which doesn't exist as L.H.L. \[=-\sqrt{2}\] whereas R.H.L. \[=\sqrt{2}\]
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