Let P = _ x → 0 ^ + (1 + ^ 2 x )(1)/(2x) , then logP is equal to

Let $P = \left(1 + \tan^{2}\sqrt{x}\right)\dfrac{1}{2x} ( {1}^{\infty}$ form$)$

If $l = \lim_{x \to a}{\left[f\left(x\right)\right]}^{g\left(x\right)},$ where $\lim_{x \to a}f\left(x\right) \to 1$ and $\lim_{x \to a}g\left(x\right) \to \infty ,$ then

$l = {e}^{\lim_{x \to a}\left[f\left(x\right) - 1\right] \cdot g\left(x\right)}$

$⇒ P = {e}^{\lim_{x \to {0}^{+}}\left(1 + \tan^{2}\sqrt{x} - 1\right) \cdot \dfrac{1}{2x}}$

$⇒ P = {e}^{\lim_{x \to {0}^{+}}\dfrac{1}{2}{\left(\dfrac{tan\sqrt{x}}{\sqrt{x}}\right)}^{2}}$

Using $\lim_{x \to 0}\left(\dfrac{\tan x}{x}\right) = 0,$ we get

$P = e\dfrac{1}{2}$

$⇒ logP = \log \left(e\dfrac{1}{2}\right) = \dfrac{1}{2}$.