```Date: Dec 4, 2012 4:47 PM
Author: Virgil
Subject: Re: Ross' Delusions re his EF.

In article <fca2fc08-f251-436c-9294-de17b9b43872@g7g2000pbi.googlegroups.com>, "Ross A. Finlayson" <ross.finlayson@gmail.com> 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|) = 1lim_(x -> 0) H(-|x|) = -1H(0) = 0But for such a function to be continuous at 0 one needslim_(x -> 0) H(|x|) = lim_(x -> 0) H(-|x|) = H(0)Or is the requirement of continuity at x = 0 different in Rossiana?--
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