bound for $|x|$ such that $|\operatorname{erf}(x)-1|<\varepsilon$?

Question

From the plot of the error function

we see that the function is close to $\pm1$ whenever $|x|\ge\frac32$ and "almost" $1$ when $|x|\ge2$. Can we give a (sharp) bound for $|x|$ such that $|\operatorname{erf}(x)-1|<\varepsilon$?

Answer 1

$\def\erf{\operatorname{erf}} \def\erfc{\operatorname{erfc}} $

As Claude Leibovici suggested, here is the simple bound.

$\begin{array}\\ \erfc(x) &=1-\erf(x)\\ &=\dfrac{2}{\sqrt{\pi}}\int_x^{\infty} e^{-t^2}dt\\ &\lt\dfrac{2}{\sqrt{\pi}}\int_x^{\infty} e^{-xt}dt \qquad t^2>tx, -t^2<-xt\\ &=\dfrac{2}{\sqrt{\pi}}\int_{x^2}^{\infty} e^{-y}\frac{dy}{x} \qquad y=xt, dy=xdt, dt=\frac{dy}{x}\\ &=\dfrac{2}{x\sqrt{\pi}}\int_{x^2}^{\infty} e^{-y}dy\\ &=\dfrac{2}{xe^{x^2}\sqrt{\pi}}\\ \end{array} $

This is the first term of the asymptotic series for $\erfc(x)$.

So to have $1-\erf(x) < c$ we can use $\dfrac{2}{xe^{x^2}\sqrt{\pi}}\le c $ or $xe^{x^2} \ge \dfrac{2}{c\sqrt{\pi}} $.

This can be solved numerically.