Let $p = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}}$ then $log$ $p$ is equal to : - DASH

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Question

Let $p = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}}$ then $log$ $p$ is equal to :

${1 \over 2}$

${1 \over 4}$

$2$

$1$

Solution

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