设函数 $f(x)=\iint_{u^2+v^2 \leqslant x^2} \arctan \left(1+\sqrt{u^2+v^2}\right) \mathrm{d} u \mathrm{~d} v(x>0)$, 则 $\lim _{x \rightarrow 0^{+}} \frac{f(x)}{\mathrm{e}^{-2 x}-1+2 x}=$
$\text{A.}$ $-\frac{\pi^2}{8}$.
$\text{B.}$ $-\frac{\pi^2}{4}$.
$\text{C.}$ $\frac{\pi^2}{4}$.
$\text{D.}$ $\frac{\pi^2}{8}$.