find smallest $x>0$ such that $\frac{A}{cx}e^{-cx^2}\le \varepsilon$ I was estimating some error and I got
$$\varepsilon(x)\le\frac{A}{cx}e^{-cx^2}$$
$A,c$ are known and positive, $x$ is also positive. 
The bigger the $x$ smaller the error. 
But I need to find the smallest possible $x$ such that $\,\varepsilon(x)\le \varepsilon$, lets say $\varepsilon = 10^{-5} $.
To be specific, I need 
$$ \inf_{x\in A}A=\left\{ \ x>0\ \ \Big\rvert \ \  \frac{A}{cx}\, e^{-cx^2}\le \varepsilon \ \right\}.$$
It does not need to be exactly infimum but try to get as close as you can.

A handy fact for this kind of problem is that
$$ f(x) = x\ln x \quad\text{and}\quad g(x) = \frac{x}{\ln x} $$
are approximately inverses, in the sense that
$$ f(g(x)) = x(1-o(1)) \quad\text{and}\quad g(f(x)) = x(1-o(1)) $$
(as $x\to\infty$).
Squaring your desired inequality and setting $u=2cx^2$, it becomes equivalent to
$$ \ln\frac{2A^2}{c\varepsilon^2} \le u\ln u $$
Applying $g$ to both sides gives that
$$ \frac{\ln\frac{2A^2}{c\epsilon^2}}{\ln\ln\frac{2A^2}{c\varepsilon^2}} \le u $$
is approximately equivalent.  If you're not too picky, at this point you might simplify by saying that $\ln\frac{2A^2}{c\varepsilon^2}\le u$ would suffice; returning to $x$ will give an estimate similar to the one you gave in the comments.
(If you are too picky to throw away the denominator there, then I think you'll also need to be more precise about the $o(1)$ in the equations for $f$ and $g$.)
