A) 1
B) \[-1\]
C) 0
D) None of these
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{\frac{1}{2}(1-\cos 2x)}}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{\frac{1}{2}.2{{\sin }^{2}}x}}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\pm \frac{\sin x}{x}\] Taking \[(+)sign,\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1\] Taking \[(-)sign,\underset{x\to 0}{\mathop{\lim }}\,\frac{-\sin x}{x}=-1\] \[\because \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{-\sin x}{x}=-1\] \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,\]does not exist.
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