Let [p = _x 0^ + ( 1 + ^2 x )^1 2x] then log p is equal to :

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 :

Accepted Answer

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

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

Applying $L$ Hospital's rule :

$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\sec \sqrt x \tan \sqrt x } \over {\sec \sqrt x .2\sqrt x }}$$

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

$$ = {1 \over 2}$$

Solutions · 1