Verified Solution • 2025–26 Session moderate

QJEEM16Ma9fa3: JEE Main 2016 Mathematics - Class 12 JEEM16Ma9fa3 Solution: Let then is equal to :

Let p=lim⁡x→0+(1+tan⁡2x)12xp = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}} then loglog pp is equal to :

Multiple Choice Options

  • A 12{1 \over 2}
  • B 14{1 \over 4}
  • C 22
  • D 11

Step-by-Step Verified Solution

ln⁡ P=lim⁡x→0+12xln⁡(1+tan⁡2x)\ln \,P = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {2x}}\ln \left( {1 + {{\tan }^2}\sqrt x } \right)

lim⁡x→0+1xln⁡(sec⁡x      )\mathop {\lim }\limits_{x \to {0^ + }} {1 \over x}\ln \left( {\sec \sqrt x \,\,\,\,\,\,} \right)

Applying LL Hospital's rule :

=lim⁡x→0+sec⁡xtan⁡xsec⁡x.2x= \mathop {\lim }\limits_{x \to {0^ + }} {{\sec \sqrt x \tan \sqrt x } \over {\sec \sqrt x .2\sqrt x }}

=lim⁡x→0+tan⁡x2x= \mathop {\lim }\limits_{x \to {0^ + }} {{\tan \sqrt x } \over {2\sqrt x }}

=12= {1 \over 2}

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