In article <email@example.com>, "Ross A. Finlayson" <firstname.lastname@example.org> wrote:
> Heh, got your goat, there, eh, Hancher. Huff and puff, you. Quit > with your ad hominem attacks.
What you have called my ad hominem, others recognise as mathematics. And much of what you have claimed as mathematics is nonsense.
> And Heaviside's step is continuous, now.
While the Heaviside's step function may be represented as the limit of a sequence of continuous Functions, in order to prove its continuity by such a limit, it would have to be a UNIFORM limit of those continuous functions, which it never can be.
Not that for the Heaviside step function H(x), lim_(x -> 0) H(|x|) = 1 lim_(x -> 0) H(-|x|) = -1 H(0) = 0
But for such a function to be continuous at 0 one needs lim_(x -> 0) H(|x|) = lim_(x -> 0) H(-|x|) = H(0)
Or is the requirement of continuity at x = 0 different in Rossiana? --