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.

Raise to the $-2$ power:
$$
\begin{align}
\frac1{\varepsilon^2}
&\le\left(\frac{A}{cx}e^{-cx^2}\right)^{-2}\\
&=\frac{c}{2A^2}2cx^2e^{2cx^2}\\
\frac{2A^2}{c\varepsilon^2}
&\le2cx^2e^{2cx^2}
\end{align}
$$
Thus,
$$
\operatorname{W}\left(\frac{2A^2}{c\varepsilon^2}\right)\le2cx^2
$$
Therefore, the smallest $x$ would be
$$
\sqrt{\frac1{2c}\operatorname{W}\left(\frac{2A^2}{c\varepsilon^2}\right)}
$$
where $\operatorname{W}$ is the Lambert W function.

For $x\ge1$, $xe^x\ge e^x$. Therefore, $x\ge\operatorname{W}(e^x)$. Thus, for $x\ge e$, $\log(x)\ge\operatorname{W}(x)$.
Therefore, if $\varepsilon\le A\sqrt{\frac2{ce}}$, then
$$
x=\sqrt{\frac1{2c}\log\left(\frac{2A^2}{c\varepsilon^2}\right)}
$$
is a decent overestimate for the $x$ necessary.
