Let $\mathrm{A}=\lim _{\mathrm{x} \rightarrow 0^{+}}\left(1+\tan ^2 \sqrt{x}\right)^{\frac{1}{2 x}}$, then $\log _{\mathrm{e}} \mathrm{A}=$

Given the limit $A = \lim{x \to 0^{+}} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}}$, which takes the indeterminate form $1^\infty$.

Apply the exponential transformation for limits of the form $[f(x)]^{g(x)}$ with $f(x) \to 1$ and $g(x) \to \infty$: $A = e^{\lim{x \to 0^{+}} g(x) [f(x) - 1]}$.

Substitute $f(x) = 1 + \tan^2 \sqrt{x}$ and $g(x) = \frac{1}{2x}$ to obtain:

$A = e^{\lim{x \to 0^{+}} \frac{\tan^2 \sqrt{x}}{2x}}$

Evaluate the exponent $L = \lim{x \to 0^{+}} \frac{\tan^2 \sqrt{x}}{2x}$ by letting $y = \sqrt{x}$, so as $x \to 0^{+}$, $y \to 0^{+}$:

$L = \frac{1}{2} \lim{y \to 0^{+}} \left( \frac{\tan y}{y} \right)^2 = \frac{1}{2} (1)^2 = \frac{1}{2}$

Thus, $A = e^{\frac{1}{2}}$. Compute $\log_e A = \log_e (e^{\frac{1}{2}}) = \frac{1}{2}$.

Final answer: $\frac{1}{2}$