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.
The (complementary) error function has a well known continued fraction expansion:
$$ \frac{1}{\sqrt{2\pi}}\int_{z}^{+\infty}e^{-x^2/2}\,dx = \frac{e^{-z^2/2}}{\sqrt{2\pi}}\cdot\frac{1}{z+\frac{1}{z+\frac{2}{z+\frac{3}{z+\ldots}}}}\tag{1}$$ 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.