A) 1
B) \[-1\]
C) 0
D) None of these
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{\left[ \frac{1}{2}(1-\cos 2x) \right]}}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{\sin }^{2}}x}}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{|\sin x|}{x}\] \[\therefore \] \[LHL=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\sin x}{-x}=-1\] And \[RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{|\sin x|}{x}=1\] Hence, limit does not exist.
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