If we define the Heaviside step function H(x) in limit notation, as per below, this yields 1/2 at x=0. How might this be adjusted to give 0 or 1 at x=0?
$$ H(x)=\lim_{b \to \infty} \frac{1}{1+\frac{1}{b}^\frac{bx}{\ln(b)}} $$
Sorry, I don't like the title but couldn't think of anything better.
You can set
$$H(x):=\lim_{b\to+\infty}\frac{\frac{2}{\pi}\arctan(bx)+1-\exp(-bx^2)}{2}$$
and this is such that $H(0)=0$, but you can modify it to cover the other case.