We obviously know the following to be equal:
$$\lim_{z \to \pm\infty} \text{erf}(z) = \pm1$$
But as far as I'm aware, the only way we know how to calculate any given value of the error function is to approximate it. How do we know the limits at infinities without any way to evaluate it algabraically?
$$\lim_{x\to \pm \infty}\text{erf}(x)=\lim_{x\to \pm \infty}\frac{2}{\sqrt \pi}\int_0^x e^{-t^2}dt = \frac{2}{\sqrt \pi} \int_0^{\pm\infty}e^{-t^2}$$ using the fact $e^{-t^2}$ is an even function: $$ \pm \frac{2}{\sqrt \pi} \int_0^\infty e^{-t^2}$$ The integral is known as the gaussian integral(half of it) and is known to equal $\frac{\sqrt \pi}{2}$ and thus, $$\lim_{z \to \pm\infty} \text{erf}(z) = \pm1$$