The unit-height impulse function can be defined as
$ I(x)=\begin{cases} 1,& \text{if } x = 0 \\ 0, & \text{otherwise} \end{cases}$
We can define the Heaviside Step Function as
$ H(x)=\begin{cases} 0,& \text{if } x < 0 \\ 1, & \text{if } x > 0 \end{cases}$
(There are variants to this $H(x)$, such as the right-continuous Heaviside Step Function, which may possibly be more appropriate to choose for the question below.)
I am wondering, is it correct to say:
$I(x) = \lim\limits_{\Delta \to 0}(H(x- \Delta) - H(x+\Delta))$
or does
$\lim\limits_{\Delta \to 0}(H(x- \Delta) - H(x+\Delta)) = 0 \text{, for all } x$, including $x=0$?
Note, I am not talking about inside an integral, where $I(x)$ is equivalent to the function that is equal to zero for all $x$
The relation of $I(x) = \lim\limits_{\Delta \to 0}(H(x- \Delta) - H(x+\Delta))$ must in fact be$$I(x) = \lim\limits_{\Delta \to 0^-}(H(x- \Delta) - H(x+\Delta))$$this is because $H(-\Delta)=1+H(\Delta)=0$ for $\Delta<0$ and as long as $\Delta\to 0^-$ we have $\Delta\ne 0$.
For $x\ne 0$, we have $$(x-\Delta)(x+\Delta)=x^2-\Delta^2>0$$ for sufficiently small $|\Delta|$. So what does this lead to?