Bound difference of gaussians

Question

For $a\geq0$ and $x>0$ let us define $$f_a(x) = \exp(-(x-a)^2) - \exp(-(x+a)^2)$$ Clearly $f_a(x) \geq 0$. I would like to find a bounding function $g(a)$ $$ 0 \leq f_a(x) \leq g(a) \quad \forall x>0 $$ With the property that $$ \lim_{a \rightarrow 0} g(a) = 0 $$

Answer 1

$$f_a(x) = \int_{a+x}^{a-x}2t e^{-t^2}\,dt\\=\int_{-x}^{x}2(t+a) e^{-t^2-a^2}e^{2at}\,dt\\=4e^{-a^2}\int_{0}^{x}(t\sinh(2at)+a \cosh(2at)) e^{-t^2}\,dt \\\leq 4e^{-a^2}\int_{0}^{x}(2at^2+a)e^{2at}\,dt. $$