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$$\mathop {\lim }\limits_{x \to 0} {{\sqrt {1 - \cos 2x} } \over {\sqrt 2 x}}$$ is

JEE · Math · previous-year question

  1. A.$$1$$
  2. B.$$-1$$
  3. C.zero
  4. D.does not existcorrect

Answer

D. does not exist

Explanation

$$\lim {{\sqrt {1 - \cos \,2x} } \over {\sqrt 2 x}} \Rightarrow \lim {{\sqrt {1 - \left( {1 - 2\,{{\sin }^2}\,x} \right)} } \over {\sqrt 2 x}}$$ $$\mathop {\lim }\limits_{x \to 0} {{\sqrt {2\,{{\sin }^2}\,x} } \over {\sqrt {2x} }} \Rightarrow \mathop {\lim }\limits_{x \to 0} {{\left| {\sin x} \right|} \over x}$$ The limit of above does not exist as $$LHS = - 1 \ne RHL = 1$$

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