How to compute this limit involving complementary error functions I am trying to take the following limit
$$\lim_{x\to \infty } \, \frac{2 x \operatorname{erfc}\left[\frac{x}{\sqrt{2} t}\right]}{t \operatorname{erfc}\left[-\frac{x}{\sqrt{2}}\right]}$$
my first thoughts were to use LHospital's rule after making the top complementary error function as the denominator of the denominator. Upon differentiating I got the following
$$\lim_{x\to \infty } \, \frac{\sqrt{2 \pi } \exp\left[\frac{\left(t^2+1\right) x^2}{2 t^2}\right] \operatorname{erfc}\left[\frac{x}{\sqrt{2} t}\right]^2}{\exp\left[\frac{x^2}{2}\right] \left(\operatorname{erf}\left[\frac{x}{\sqrt{2}}\right]+1\right)+t \exp\left[\frac{x^2}{2 t^2} \right]\operatorname{erfc}\left[\frac{x}{\sqrt{2} t}\right]}$$
which seems even more complicated.
Maybe using an identity or some form of expansion for the complementary error function may lead to a simpler expression.
Any help will be greatly appreciated.
Thank you.

If you look here and here, you will see the asymptotic expansion
$$\operatorname{erfc}(z) = \frac{e^{-z^2}}{z\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n - 1)!!}{(2z^2)^n}$$ $$\operatorname{erfc}(-z) = 2-\frac{e^{-z^2}}{z\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n - 1)!!}{(2z^2)^n}$$So, for very large $z$, keeping the first term only  makes
$$\operatorname{erfc}\left(\frac x {t \sqrt 2}\right) \approx\frac{\sqrt{\frac{2}{\pi }} t e^{-\frac{x^2}{2 t^2}}}{x}$$ $$\operatorname{erfc}\left(-\frac x { \sqrt 2}\right) \approx 2-\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2}}}{x}$$
$$\frac{2 x \operatorname{erfc}\left(\frac{x}{\sqrt{2} t}\right)}{t \operatorname{erfc}\left(-\frac{x}{\sqrt{2}}\right)}\approx \frac{2 \sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2 t^2}}}{2-\frac{\sqrt{\frac{2}{\pi }}
   e^{-\frac{x^2}{2}}}{x}} \to \sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2 t^2}}$$
