QJEEM11Ma5f57: JEE Main 2011 Mathematics - Class 12 JEEM11Ma5f57 Solution

Question

Verified Solution • 2025–26 Session moderate

\mathop {\lim }\limits_{x \to 2} \left( {{{\sqrt {1 - \cos \left\{ {2(x - 2)} \right\}} } \over {x - 2}}} \right)

Accepted Answer

\mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}

= \mathop {\lim }\limits_{x \to 2} {{\sqrt 2 \left| {\sin \left( {x - 2} \right)} \right|} \over {x - 2}}

L.H.L. = \mathop {\lim }\limits_{x \to {2^ - }} {{\sqrt 2 \sin \left( {x - 2} \right)} \over {\left( {x - 2} \right)}} = - \sqrt 2

R.H.L. = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {x - 2} \right)} \over {\left( {x - 2} \right)}} = \sqrt 2

Thus

L.H.L. \ne R.H.L.

Hence,

\mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}\,\,

does not exist.

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