# Is the $ϵ,δ$ definition of a limit not well-defined?

He lays out his objections against the $ϵ,δ$ definition around 14 min.

Here is the discription of the video:

In this video we aim to give a precise and simpler definition for what it means to say that: a rational polynumber on-sequence p(n) has a limit A, for some rational number A. Our definition is both much simpler and more logical than the usual epsilon -delta definition found in calculus texts. What is required is that we need to find two natural numbers: k called the scale, and m called the start that allow us to bound in a pretty simple way the difference between p(n) and A.

The epsilon-delta definition of a limit is usually considered a high point of logical rigour. Not so. It is also considered too logically involving to be taken seriously as a pedagogical pillar for most undergrads. Hence students may be told about the definition, but are not required to seriously understand it, or be able to use it--unless they are prospective maths majors.

There is a subtle ambiguity in the definition: given an epsilon we are supposed to demonstrate there is a delta (with certain properties) but how are we to do this, since an potential infinity of epsilons are involved? In practice what is required is a correspondence (function/relation etc) between epsilon and delta but the nature of this required correspondence is not clear. We return to our familiar conundrum of using the workfunction'' without a proper definition of it.

The key point that makes our simpler more intuitive notion of limit of a sequence work is that we are dealing with very particular and clearly defined on-sequences: those generated by a rationl polynumber. A good example of the benefits of being careful rather than casual when dealing with the foundations of analysis!

My question is: Is this an opinion shared by more mathematicians ? I kind a feel like that this Professor of the University of New South Wales is standing completely alone as it comes to this. I don't really undestand his objections, but I don't think I'm skilled enough to understand if his objections are legit.

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There is no ambiguity in $\epsilon-\delta$. –  Thomas Andrews Apr 18 '13 at 12:51
He seems to think that we have to come up with an assignment $\varepsilon \mapsto \delta$... not so. –  Clive Newstead Apr 18 '13 at 12:55
And with the axiom of choice, such a function exists. –  Karl Kronenfeld Apr 18 '13 at 12:55
It is also considered too logically involving to be taken seriously as a pedagogical pillar for most undergrads. It's hard not to read this as "undergrads are too dumb to understand that." I think that is manifestly wrong: they can understand it. The definition is both transparent and accessible. I do not claim that it is obvious or that an untrained person should understand it immediately, I just mean that in a reasonable amount of time anyone can be convinced it captures the intuitive meaning. I would take things this prof says with a grain of salt. –  rschwieb Apr 18 '13 at 13:12
There seems to be a discussion at xkcd about this individual, if I have not made some gross mistake. –  rschwieb Apr 18 '13 at 13:19