Is there any way to find the supremum of $\lim_{x \to 0} \sup_x \frac{{\operatorname{erfc}(x)\log (\operatorname{erfc}(x))}}{{{x^2}}}$

Question

My main objective is to find the supremum of a function$\def\erfc{\operatorname{erfc}}$

$$\sup_x \frac{{\erfc(x)\log (\erfc(x))}}{x^2}$$

My question is can there be any supremum at $x=0$ even though $\frac{{\erfc(x)\log (\erfc(x))}}{x^2}$ is undefined at this point. Can I use l'Hospitale rule to take the derivative above and below to find the

$$\sup_x \frac{{\erfc(x)\log (\erfc(x))}}{x^2}$$

Thank you.

Answer 1

To see what happens close to $x=0$, compose Taylor series $$\text{erfc}(x)=1-\frac{2 x}{\sqrt{\pi }}+\frac{2 x^3}{3 \sqrt{\pi }}+O\left(x^4\right)$$ $$\log (\text{erfc}(x))=-\frac{2 x}{\sqrt{\pi }}-\frac{2 x^2}{\pi }+\frac{1}{3} \left(\frac{2}{\sqrt{\pi }}-\frac{8}{\pi ^{3/2}}\right) x^3+O\left(x^4\right)$$

$$\frac{\text{erfc}(x)\log (\text{erfc}(x))} {x^2}=-\frac{2}{x\sqrt{\pi }}+\frac{2}{\pi }+\frac{2 (2+\pi ) x}{3 \pi ^{3/2}}+O\left(x^2\right)$$