# What's an expression for the function of a limit?

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.

• Why don’t you simply subtract/add 1/2? May 20, 2021 at 15:20
• Adding or subtracting a half would cause the whole function to shift up or down, but I'm looking for an expression that shifts it up or down at just the point x=0. May 20, 2021 at 18:03
• Why do you need to define $H(x)$ as a limit? What's wrong with simply writing a piecewise definition? May 21, 2021 at 5:38
• I'm working on a math that allows for defining differentiable step functions using novel numbers and was wondering if it has any utility in intuiting solutions that may otherwise not be obvious. All its expressions can be expressed as limits to zero or infinity, so I figured if it has any utility, it would be something about the notation itself. May 21, 2021 at 11:21

$$H(x):=\lim_{b\to+\infty}\frac{\frac{2}{\pi}\arctan(bx)+1-\exp(-bx^2)}{2}$$
and this is such that $$H(0)=0$$, but you can modify it to cover the other case.
• The idea is the same as in your example: let me talk about that. You have that your limit is $1/2$ when computed for $x=0$. It means you just need to subtract a functions which is $1/2$ in $x=0$ and $0$ everywhere else. Clearly $1/2\exp[-bx^2]$ in the limit does the job. This is to say that you can also adjust your example. I thought of using the arctangent because in the limit it becomes the step function (almost). Then I adjusted from there Apr 17, 2022 at 7:39