MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $
  • A
    $1$
  • B
    $-1$
  • C
    $0$
  • None of these

Answer

Correct option: D.
None of these
d
(d) $\mathop {\lim }\limits_{x \to 0} \,\frac{{\sqrt {{\textstyle{1 \over 2}}(1 - \cos 2x)} }}{x} = \mathop {\lim }\limits_{x \to 0} \,\frac{{|\,\,\sin x\,\,|}}{x}$

So, $\mathop {\lim }\limits_{x \to 0 + } \,\frac{{|\,\sin x\,|}}{x} = 1$ and $\mathop {\lim }\limits_{x \to 0 - } \,\frac{{|\,\sin x\,|}}{x} = - 1$

Hence limit does not exist.

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