A) 0
B) 1
C) \[\sqrt{2}\]
D) does not exist
Correct Answer: D
Solution :
We have, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos x}}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{2}\left| \sin \frac{x}{2} \right|}{x}\] \[=\sqrt{2}\underset{x\to 0}{\mathop{\lim }}\,\frac{\left| \sin \frac{x}{2} \right|}{x}\] It can be easily checked that LHL at \[x=0\]is\[x=-\frac{1}{\sqrt{2}}\]and RHL, at \[x=0\]is \[\frac{1}{\sqrt{2}}.\] Hence, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos x}}{x}\]does not exist.
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