- #36

DeaconJohn

- 122

- 0

**delta epsilon explanation**

I can't get my head around the epsilon-delta definition of a limit. Unfortunately I don't have a teacher to ask (I'm teaching this to myself as a self interest) so this forum is my last resort -- google hasn't been kind to me.

From what I've seen, I don't really understand how the definition means much of anything (visual examples included.) All it seems to say is "there is an unspecified value which is greater than the difference between a function and a given value L, where that difference is greater than zero" The problem is, I don't see any need for L or f(x) to be anywhere near one another.

There are three things I would like to say. First, your concern over your lack of understanding of the delta epsilon definition of a limit shows that you have significant mathematical talent.

Second, here is the way I like to phrase the delta-epsilon definition:

DEFINITION: "A function f:R->R aproaches a limit L as x goes to c if and only if

For all epsilon > 0, there exists a delta > 0 such that |x-c| < delta implies that |f(x)-L| < epsilon."

Note that I didn't say anything about "an unspecified value of epsilon." I said, "for ALL positive values of epsilon.

It is an implication of the definition that it is only small values of epsilon that really count. This is because if you can find a "delta," call it delta1, for some small value of epsilon, call it epsilon1, then that value of delta will work for all epsilon greater than epsilon1.

In other words, if you are trying to apply the definition to prove that some number L is a limit as x -> c of some function f(x) and you suceed in finding a delta (delta1) that works for some value of epsilon (epsilon1), then that value of delta (delta1) will work for all epsilon2 > epsilon1. Hence you are finished with values of epsilon greater than epsilon1 and all you have to worry about now is the values of epsilon less than epsilon1.

One could state this as a Theorem:

Suppose that there exists epsilon1, delta1 > zero such that |f(x)-L| < epsilon1 for all x such that |x-c| < delta1, THEN for all epsilon2 > epsilon1 it is true that |f(x)-L| < epsilon2.

Proof: |f(x)-L| < epsilon1 < epsilon2. QED.

The third thing I would like to say is that for the three months of summer vacation before I went to college 45 years ago, I spent a lot of time trying to understand the delta-epsilon definition of a limit and of a derivative. I didn't feel I understood it after those three months. But, a couple of weeks into my freshman calculus class, I felt I really understood it. I don't remember much else about what happened back then, but, I do remember that the Professor invited me to switch from the regular calculus class to the honors calculus class after the first month of class. I thanked him and said that I was learning too much in the regular class to switch, but, could I take both? He said "of course." The point is that the manyt many hours over many many months of thinking about that definition served me well not only as an undergraduate but also as a graduate student. All phases of analysis (indlucing PDE, functional analysis, and C* algebras) were vastly much easier for me because I learned that definition. Same thing with mathematical logic and general topology.

Algebra and algebraic topology and algebraic geometry was difficult for me, though, and still is. Not to mention number theory. I still don't have my belt all the way around Chern classes.

Deacon John