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

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

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

$\Rightarrow P={\mathrm{e}}^{\underset{x\to 0^{+}}{\lim }\left(1+{\tan }^2\sqrt{x}-1\right)·\frac{1}{2x}}$

$\Rightarrow P={\mathrm{e}}^{\underset{x\to 0^{+}}{\lim }\frac{1}{2}{\left(\frac{\tan \sqrt{x}}{\sqrt{x}}\right)}^2}$

Using $\underset{x\to 0}{\lim }\left(\frac{\tan x}{x}\right)=0,$ we get

$P={\mathrm{e}}^{\frac{1}{2}}$

$\Rightarrow \log P=\log \left(e^{\frac{1}{2}}\right)=\frac{1}{2}$.