Show that $\frac1{\sqrt a}e^{-\frac{b^2}{8a}}$ is bounded

Question

I want to show that $f(a,b):=\frac1{\sqrt a}e^{-\frac{b^2}{8a}}$ for $a>0$ and $b\in\mathbb R\setminus\{0\}$ is bounded.

Note that for fixed $b\in\mathbb R\setminus\{0\}$, $f(a,b)\to0$ as $a\to0$ and as $a\to\infty$. So, it's sufficient to determine the local extrema of $f$.

Noting that $$\frac\partial{\partial a}f(a,b)=0\Leftrightarrow a=\frac{b^2}4$$ and hence there is a unique local extremum at $a=\frac{b^2}4$, which is a local maximum since $\frac{\partial^2}{\partial a^2}f(a,b)<0$ at $a=\frac{b^2}4$.

Together, this should yield that $f(a,b)\in\left(0,\frac2{|b|\sqrt e}\right]$ for all $a>0$ and $b\in\mathbb R$. But since $\frac2{|b|\sqrt e}\xrightarrow{|b|\to0}\infty$, this is not sufficient.

Obviously, $f(a,b)\le1$ if $a\ge1$, so we only need to consider $a\in(0,1)$.

So, what do we need to do?

Note that the desired claim is made in the following answer (but I don't understand the argument): https://math.stackexchange.com/a/3203657/47771.

Answer 1

Not true. $f(a,0) = \frac{1}{\sqrt{a}} \to \infty$ as $a \to 0+$.