A) 1
B) \[-1\]
C) 0
D) None of these
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\sqrt{\frac{\frac{1-\cos 2x}{2}}{x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{\frac{1-1+2{{\sin }^{2}}x}{2}}}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{|\sin x|}{x}\] \[LHL=\underset{x\to 0}{\mathop{\lim }}\,-\frac{\sin x}{x}=-1\] and \[RHL=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1\] Hence, limit does not exist.You need to login to perform this action.
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