Q: Asymptotic rate of decrease of error function The complementary error function is defined as
$$
\text{erfc}(x) = 1 - \frac{2}{\sqrt{\pi}}\int_0^{x} e^{-t^2} dt
$$
and is related to the Gaussian (Normal) distribution. Is there an approximation of the form $\exp(g(x))$ that converges to $\phi(x)$ asymptotically? i.e. can we find $g(x)$ such that
$$
\underset{x \rightarrow \infty}{\lim} \frac{\exp(g(x))}{\text{erfc}(x)} = 1
$$
and that $\exp(g(x))$ "approximates" $\text{erfc}(x)$ in some sense when $x$ is large but not infinite? 

A: There is one asymptotic expansion in the Handbook of Mathematical functions and also listed in your link; it is not quite in the form you wanted but close.
If 
$$\text{erfc}(z) = \frac{2}{\sqrt{\pi}} \int_z^\infty e^{-t^2}~ dt = 1 - \text{erf}(z)$$
then
$$\text{erfc}(z)\sim \frac{1}{\sqrt{\pi}}e^{-z^2} \cdot 
\sum_{k=0}^\infty 
 (-1)^k~\frac{(2k)!}{2^{2k}k! } \cdot \frac{1}{z^{2k+1}}
$$
as $z\to\infty$, $\lvert \arg z \rvert < 3\pi/4. $
This can be adapted to the required form;  for example, the first term is,
$$\text{erfc}(z) = \frac{1}{\sqrt \pi} e^{-z^2}\cdot \frac{1}{z} = e^{-z^2-\log(z\sqrt{\pi})}$$
which gives $g(z) = -z^2-\log(z\sqrt{\pi})$.
