Proving approximation of $\text{erf}$ with Taylor expansion I am asked to show that 
$$\text{erf}(x) \approx 1 - \frac{1}{\sqrt{\pi}}\frac{1}{x}e^{-x^2}$$
in a computational project. Numerically it is really easy to show that this approximation makes sense. I want, however, to look for the process that allowed someone to come up with this approximation.
For that matter, I assumed maybe one could do something like a Taylor expansion "around infinity".
I typed
Series[Erf[x], {x, Infinity, 1}]
in mathematica and it gave me this exact approximation. I browsed online for how to "expand a function around $\infty$" and assuming everything works (i.e. without worrying about the details that let me write a given expression and whatnot)
I defined $$f(x) = \text{erf}\left(\frac{1}{x}\right) = \frac{2}{\sqrt{\pi}}\int_0^{\frac{1}{x}} e^{-\xi^2}d\xi$$
and then decided to expand $f$ around $\epsilon$ where we would have
$$f(\epsilon) \approx f(x) + (\epsilon - x)f'(x)$$
and then I computed $f'(x)$:
$$f'(x) = \frac{2}{\sqrt{\pi}} \left(\int_0^{\frac{1}{x}} e^{-\xi^2}d\xi \right)' = \frac{2}{\sqrt{\pi}}\left(\frac{1}{x} \right)' e^{-\frac{1}{x^2}} = -\frac{2}{\sqrt{\pi}}\frac{1}{x^2}e^{-\frac{1}{x^2}}$$
plugging that in the first expression and letting $\epsilon \to 0$ gives
$$1 \approx \text{erf}\left(\frac{1}{x}\right) + \frac{2}{\sqrt{\pi}}\frac{1}{x}e^{-\frac{1}{x^2}}$$
then writing $y = 1/x$ gives $$\text{erf}(y) \approx 1 - \frac{2}{\sqrt{\pi}}ye^{-y^2}$$ 
which is close, but not quite what I am looking for... Where did I make a mistake??
 A: The reason for expecting such approximation is pretty simple: the derivative of $\text{erfc}(x)=\frac{2}{\sqrt{\pi}}\int_{x}^{+\infty}e^{-t^2}\,dt$ is $-\frac{2}{\sqrt{\pi}}e^{-x^2}$, hence by De l'Hopital rule
$$ \lim_{x\to +\infty}\frac{\text{erfc}(x)}{\frac{1}{x\sqrt{\pi}}e^{-x^2}}=\lim_{x\to +\infty}\frac{-\frac{2}{\sqrt{\pi}}e^{-x^2}}{-\frac{2}{\sqrt{\pi}}e^{-x^2}+\frac{1}{x^2\sqrt{\pi}}e^{-x^2}}=\lim_{x\to +\infty}\frac{1}{1-\frac{\sqrt{\pi}}{2x^2}}=1$$
and $\text{erfc}(x)\sim \frac{1}{x\sqrt{\pi}}e^{-x^2}$ as $x\to \infty$. Actually we have
$$\text{erfc}(x)=\frac{e^{-x^2}}{\sqrt{\pi}}\cdot\frac{1}{x+\frac{1/2}{x+\frac{1}{x+\frac{3/2}{x+\ldots}}}}$$
for any $x\gg 0$.
A: We can write:
$$\operatorname{erfc}(x) = \frac{2}{\sqrt{\pi}}\int_x^\infty\exp\left(-t^2\right)dt =  \frac{1}{\sqrt{\pi}}\int_{x^2}^\infty\frac{\exp\left(-u\right)}{\sqrt{u}}du$$
Partial integration yields:
$$\operatorname{erfc}(x) = \frac{1}{\sqrt{\pi}}\frac{\exp\left(-x^2\right)}{x}-\frac{1}{2\sqrt{\pi}}\int_{x^2}^\infty\frac{\exp\left(-u\right)}{u^{3/2}}du$$
And we can go on to repeatedly perform partial integrations to get to an asymptotic series:
$$\operatorname{erfc}(x) = \frac{\exp\left(-x^2\right)}{\sqrt{\pi}}\left(\frac{1}{x}-\frac{1}{2 x^3}+\frac{3}{4 x^5}-\frac{15}{8 x^7}+\cdots\right)$$
A: From the initial formula
$$(\text{erf}(x))'\approx\frac2{\sqrt\pi}e^{-x^2}+\frac1{\sqrt\pi x^2}e^{-x^2}.$$
The second term is tiny in front of the first.
A: An alternative approach: You might find this inequality handy:
$$
\frac{x^{2}}{1+x^{2}}\frac{1}{x}\exp \left( \frac{-x^{2}}{2}\right)\leqslant \int_{x}^{\infty}e^{-t^{2}/2}dt\leqslant \frac{1}{x}\exp\left(\frac{-x^{2}}{2}\right)
$$
To apply this to your exercise you'll need to get rid of that division by 2.
