JEE MAIN - Mathematics (2016 (Offline) - No. 1)
Let $$p = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}}$$ then $$log$$ $$p$$ is equal to :
$${1 \over 2}$$
$${1 \over 4}$$
$$2$$
$$1$$
Explanation
$$\ln \,P = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {2x}}\ln \left( {1 + {{\tan }^2}\sqrt x } \right)$$
$$\mathop {\lim }\limits_{x \to {0^ + }} {1 \over x}\ln \left( {\sec \sqrt x \,\,\,\,\,\,} \right)$$
Applying $$L$$ Hospital's rule :
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\sec \sqrt x \tan \sqrt x } \over {\sec \sqrt x .2\sqrt x }}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\tan \sqrt x } \over {2\sqrt x }}$$
$$ = {1 \over 2}$$
$$\mathop {\lim }\limits_{x \to {0^ + }} {1 \over x}\ln \left( {\sec \sqrt x \,\,\,\,\,\,} \right)$$
Applying $$L$$ Hospital's rule :
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\sec \sqrt x \tan \sqrt x } \over {\sec \sqrt x .2\sqrt x }}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\tan \sqrt x } \over {2\sqrt x }}$$
$$ = {1 \over 2}$$
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