Epsilon-delta definition

What particularly are you not understanding? Do you mean the [itex]\epsilon[/itex]-[itex]\delta[/itex] definition of continuity of a function at a point?

In this case it reads:

A function [itex]f:D \rightarrow \mathbb{R}[/itex] is continuous at [itex]x_0 \in D[/itex] iff for any [itex]\epsilon>0[/itex] there exists a [itex]\delta>0[/itex] such that for all [itex]x \in D[/itex] with [itex]|x-x_0|<\delta[/itex]
[tex]|f(x)-f(x_0)|<\epsilon.[/tex]
This just says in a formal way that the graph of the function doesn't jump at [itex](x_0,f(x_0))[/itex].

 

##0<|x-x_0|<\delta## actually.

 

https://www.physicsforums.com/showthread.php?t=700787

 

There's no reason to exclude the point ##x = x_0##. We trivially have ##|f(x_0) - f(x_0)| < \epsilon## for any ##\epsilon##.

 

It sure is a reason.
Otherwise, discontinuous functions would be deprived of limit values at the point of discontinuity. Thus, the limit concept would be conflated with the continuity concept.
Think about it!

 

But vanhees71 was giving the definition of continuity at ##x_0##, not the definition of the existence of a limit at ##x_0##. If the function is to be continuous at ##x_0##, then it must be defined at ##x_0## and have the correct value!

 

Hmm..no read it again.
What he posted was the definition in terms of the LIMIT concept in which L=f(x_0).
In particular, he writes d>0

 

One more remark and then I'll shut up. :tongue: Here's a discussion from Spivak's Calculus (chapter 6, before Theorem 2) which may clarify:

 

And I do not see why we should make a special case for the continuity criterion relative to the general criterion for a limit?

Why change a perfectly good criterion for continuity?