I'm trying to find the behaviour of the error function, $erf(z)$ as $z \rightarrow -\infty$
$$erf(z) = \frac{2}{\sqrt{\pi}}\int_0^{z} e^{-s^2}\mathrm{ds}$$
I know that we can find the limit of $erf(z) \rightarrow 1$ as $z \rightarrow \infty$ by the Gaussian integral, can that result be used somehow to find the result as $z \rightarrow -\infty$?
As $z \to - \infty$, make a change of variables in $s$ to rewrite the integral - that is, $$ erf(- \infty) = \frac{2}{\sqrt{\pi}}\int_0^{ - \infty} e^{-s^2}\mathrm{ds} = - \frac{2}{\sqrt{\pi}}\int_{- \infty} ^{0} e^{-s^2}\mathrm{ds} . $$ Taking $x = -s$, we have $$ erf ( - \infty) = - \frac{2}{\sqrt{\pi}} \left( -\int_{\infty} ^{0} e^{-x^2}\mathrm{d}x \right) = - \frac{2}{\sqrt{\pi}} \int_{0} ^{\infty} e^{-x^2}\mathrm{d}x . $$ From this the answer is easy to see.