Is the error function only = 1 at infinity? I have a little question regarding the error function that disturbs me a lot:
Let us consider the classical error function defined as:
erf($z$) $= \frac{2}{\sqrt{\pi}}\int_{0}^{z} e^{-t^2} dt$
It is well-known [1] that for $z= \infty,$ erf($z$) = 1. My question may be idiotic but is the following one: is the error function only equals to 1 at $z= \infty$ ? In other words, is the error function never reaches the value of 1 before the strict value $z= \infty$ ? Don't tell me to refer to tables of values of erf($z$) because the latter are based on numerical integrations and thus subjected to numerical uncertainties.
Thank you very much for your help,
Regards
[1] http://mathworld.wolfram.com/Erf.html
 A: Probably not very useful.
In the linked page, they give the asymptotic series
$$\text{erf}(x)=1-\frac{e^{-x^2}}{\sqrt{\pi }}\sum_{n=0}^\infty (-1)^n \frac{(2n-1)!!}{2^n} x^{-(2n+1)}$$ and a very accurate asymptotics would be
$$\text{erf}(x)=1-e^{-x^2} \left(\frac{1}{\sqrt{\pi }
   x}+O\left(\frac{1}{x^2}\right)\right)$$
For $x=10$, the "exact" value is 
$$0.99999999999999999999999999999999999999999999791151$$ while the above truncated expression gives
$$0.99999999999999999999999999999999999999999999790117$$
Edit
After GNU Supporter's comment, consider
$$a_n=(-1)^n \frac{(2n-1)!!}{2^n} x^{-(2n+1)}$$ This gives
$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{2 n+1}{2 x^2}\approx \frac n {x^2}$$ which decreases very fast.
A: If you see that:
$$\frac{d}{dz} \text{erf}(z) = \frac{2e^{-z^2}}{\sqrt{\pi}}>0$$
Therefore the function is strictly increasing.


*

*As $\text{erf}(0)=0$  then $\text{erf}(x)>0$. 

*$\frac{2}{\sqrt{\pi}}\int_{0}^{\infty} e^{-t^2} dt=1$


Take $x>0$.
$$\frac{2}{\sqrt{\pi}}\int_{0}^{\infty} e^{-t^2} dt = \frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2} dt + \frac{2}{\sqrt{\pi}}\int_{x}^{\infty} 
e^{-t^2} dt $$
$$1 = \text{erf}(x) + \frac{2}{\sqrt{\pi}}\int_{x}^{\infty} e^{-t^2} dt$$
$$1- \text{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{x}^{\infty} e^{-t^2} dt >0$$
$$1>\text{erf}(x)$$
