# Epsilon-delta definition

1. Oct 15, 2013

### imram.math

please i am new to math. I donot know exact meanings of epsilon-delta definition. i dont comprehend it. Would Anybody help me. thanks in advance

2. Oct 15, 2013

### vanhees71

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

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

3. Oct 15, 2013

### pwsnafu

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

4. Oct 15, 2013

### WannabeNewton

5. Oct 15, 2013

### jbunniii

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

6. Oct 15, 2013

### arildno

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.

7. Oct 15, 2013

### jbunniii

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!

8. Oct 15, 2013

### arildno

What he posted was the definition in terms of the LIMIT concept in which L=f(x_0).
In particular, he writes d>0

9. Oct 15, 2013

### jbunniii

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:

10. Oct 15, 2013

### arildno

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?