Asymptoting to 0: is erfc(z) quicker than exp(-z)?

Question

If I have a function of the form $\mathrm{erfc}\left(z\right)/e^{-z}$, should I expect its limit at large $z$ to be $0$ or $\infty$?

My instinct is that it should be $0$, by considering the asymptotic approximation of the complimentary error function, which has a factor of $e^{-z^2}/z$. Is this true?

Answer 1

Hint: If $z>0 $ we have the bounds $$\frac{2}{\sqrt{\pi}}\frac{e^{-z^2}}{z+\sqrt{z^{2}+2}}\leq\textrm{erfc}\left(z\right)\leq\frac{2}{\sqrt{\pi}}\frac{e^{-z^2}}{z+\sqrt{z^{2}+\frac{4}{\pi}}}$$