 Logarithmic inversion of a Gaussian divided by its argument

For $ϵ\downarrow 0$, put $L=\log (1/ϵ)$. Taking a logarithm gives $x^2+\log x=L$, first implying $x\sim \sqrt{L}$. Substitution of $x=\sqrt{L}+d$ gives $2\sqrt{L}d+\frac{1}{2}\log L$ at first correction order. Hence $d=-(\log L)/(4\sqrt{L})$; substituting back and using the derivative of $x^2+\log x$ bounds the remaining root error by $O((\log L{)}^2/L^{3/2})$.