I know that $$\lim_{a \to -\infty}{\sqrt{\pi t}\left[\operatorname{erf} \left(\frac{x-a}{2 \sqrt{t}}\right) - \operatorname{erf} \left(\frac{x}{2 \sqrt{t}}\right)\right]} = \sqrt{\pi t} \operatorname{erfc} \left(\frac{x}{2 \sqrt{t}}\right)$$ where $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x}{e^{-t^2}}dt$ is the error function and $\operatorname{erfc}(x) = 1 - \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{x}^{\infty}{e^{-t^2}}dt$ is the complementary error function.
My question now is: what can I say about the next limit?
$$\lim_{b \to \infty}{\sqrt{\pi t}\left[\operatorname{erf} \left(\frac{x}{2 \sqrt{t}}\right) - \operatorname{erf} \left(\frac{x-b}{2 \sqrt{t}}\right)\right]} = \text{?}$$
You can move the $\sqrt{\pi t}$ term outside the limit to get $$\sqrt{\pi t}\lim_{b \to \infty}\left[\operatorname{erf} \left(\frac{x}{2 \sqrt{t}}\right) - \operatorname{erf} \left(\frac{x-b}{2 \sqrt{t}}\right)\right]$$
This limit can be split into two limits, the first of which is a constant value, making it $$\sqrt{\pi t}\left(\operatorname{erf} \left(\frac{x}{2 \sqrt{t}}\right)-\lim_{b \to \infty} \operatorname{erf} \left(\frac{x-b}{2 \sqrt{t}}\right)\right)$$
Finally since $\lim_{z \to -\infty}\operatorname{erf}(z) = -1$, the expression simplifies to $$\sqrt{\pi t}\left(\operatorname{erf} \left(\frac{x}{2 \sqrt{t}}\right)+1\right)$$