= Solution
{c}
The <function> $e^{-x^2}/x$ is strictly decreasing from infinity to zero on $x>0$, so there is a unique positive root, which tends to infinity. Let $L=\log(1/\epsilon)$. Taking <logarithms> gives
$$
x^2+\log x=L.
$$
First $x^2\sim L$. Put $x=\sqrt L+d$ with $d=o(\sqrt L)$, expand, and cancel $L$:
$$
2\sqrt L\,d+d^2+\tfrac12\log L
+\log(1+d/\sqrt L)=0.
$$
The leading balance gives $d=-(\log L)/(4\sqrt L)$. Substitution leaves a residual $O((\log L)^2/L)$; the <derivative> of $x^2+\log x$ is of order $\sqrt L$, so the root error is $O((\log L)^2/L^{3/2})$. Thus
$$
\boxed{x=\sqrt L-\frac{\log L}{4\sqrt L}
+O\!\left(\frac{(\log L)^2}{L^{3/2}}\right).}
$$
The second term is logarithmic rather than a constant multiple of $L^{-1/2}$. This is <logarithmic inversion of a Gaussian divided by its argument>.
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