# Good upper bound for $1-\operatorname{Erf}(x)$

For error function $\text{Erf}(x)$ I mean $$\operatorname{Erf}(x) = \int_{-\infty}^x\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}u^2\right)\mathrm{d} u.$$

My statistics professor said that

$$1-\operatorname{Erf}(x) \leq \frac{1}{x}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2} x^2\right)$$ and that $$\lim_{x\to+\infty}\frac{1-\operatorname{Erf}(x)}{\frac{1}{x}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}x^2\right)} = 1.$$

Proving the first fact is super easy, what about the second one?

EDIT: there were some mistakes, now it is all fixed.

• Usually $\operatorname{Erf}(x)$ would mean the integral of the function you've got, i.e. $$\operatorname{Erf} (x) = \int_{-\infty}^x \frac 1 {\sqrt{2\pi}} e^{-u^2/2} \,du,$$ or else a rescaled version of that. $\qquad$ Feb 4 '17 at 18:21
• You're using both $x$ and $z$ to represent something. Are they both supposed to be the same thing? Feb 4 '17 at 18:22
• Personal attempts? Explanations about why the "super easy" proof does not work to prove the second fact? Anything?
– Did
Feb 4 '17 at 18:24
• @MichaelHardy fixed. Feb 4 '17 at 18:26
• Well, using the change of variable $$u=x+\frac{v}x$$ one gets that, for every positive $x$, $\sqrt{2\pi}(1-\mathrm{erf}(x))$ equals $$\int_x^\infty e^{-u^2/2}du=\int_0^\infty e^{-x^2/2-v-v^2/(2x^2)}\frac{dv}x=\frac{e^{-x^2/2}}x\int_0^\infty e^{-v-v^2/(2x^2)}dv$$ and, when $x\to\infty$, the last integral increases to $$\int_0^\infty e^{-v}dv=1$$ as desired. (No L'Hopital, no fancy continued fraction, just a good ol' linear change of variable...)
– Did
Feb 4 '17 at 21:13

$$\frac{1}{\sqrt{2\pi}}\int_{z}^{+\infty}e^{-x^2/2}\,dx = \frac{e^{-z^2/2}}{\sqrt{2\pi}}\cdot\frac{1}{z+\frac{1}{z+\frac{2}{z+\frac{3}{z+\ldots}}}}\tag{1}$$ that is not difficult to prove by studying the recurrence relation fulfilled by the moments $$M_n = \frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty} x^n e^{-x^2/2}\,dx.\tag{2}$$ Anyway, to prove the second limit it is enough to apply de l'Hospital rule.