Question about epsilon-delta proofs in general

Question

I think I understand how the proofs work, but I'm not sure if I'm right, I have 2 thoughts:

The existence of $\delta$ implying

$ | f(x) - L | < \epsilon$

leads to the end of the proof, is it because $ | f(x) - L | < \epsilon$ implies that this absolute value is always less than a finite number and it is strictly positive, so it itself must be finite? or..

Is it because we have the freedom of choice of $\epsilon$ so we can choose it to be really really close to zero?

Also in the continuity definition, $ | f(x) - f(a) | < \epsilon$

same questions, does the proof end there because $ | f(x) - f(a) |$ is finite or is it because we can choose $\epsilon$ very small such that we esentially have $f(x) = f(a)$?

Was I clear in these questions, I looked into some epsilon delta questions in this site but I can't find the same reasoning of my question.

Answer 1

Well, ultimately the proof is because the conditions of the definition are met.

The definition of continuous is that such a $\delta$ exist so proving it does, means it is continuous. By definition.

If we defined a SCHNICKLEBACK is a toad that spits green algae when you put pink flannel on its toe, the we can prove a toad is a SCHNICKLEBACK by showing that if we always put pink flannel on its toe it will spit green algae. What is the idea behind it. Is it because the flannel is pink? Is it because the pink thing is flannel, or does it have to do with the algae. Well, none of that. A SCHNICKLEBACK is what say it is, and in this case we are just proving what we said it was.

$\lim_{x\to a} f(x)= L$ means that for every $\epsilon > 0$ there exists a $\delta >0$ so that whenever $|x-a|$ then $|f(x) - L| < \epsilon$. So if we that such a $\delta$ will exist for every $\epsilon$ then ...... such a $\delta$ will exist for every $\epsilon$. We said that whenever that happens its a SCHNICKLEBACK and it happens, so it is a SCHNICKLEBACK and... oops, I'm sorry... we said whenever that happens its a LIMITASXGOESTOAOFFOFXEQUALSL. So that happens so it is a LIMITASXGOESTOAOFFOFXEQUALSL.

A couple of points:

One for it to be a "$\lim_{x\to a}f(x) = L$" it has to be true for any $\epsilon$ no matter how small so that a $\delta$ exists. It won't do if we did it for just one specific $\epsilon$. It must be that any epsilon we choose at all, so long as it is $>0$ there will be an $\delta$ (which will very likely depend on which epsilon; a different $\epsilon$ may have a different delta but every possible $\epsilon$ will have some $\delta).

I suppose your "real question" is why do we have this definition and what does it "really mean". So why do we have the definitions of a SCHNICKLEBACK is a toad that spits green algae when you put pink flannel on its toe? Well, because I'm sitting here drunk thinking it is funny. Is the defininiton of "$\lim_{x\to a}f(x) = L$" being "that for every $\epsilon > 0$ there exists a $\delta >0$ so that whenever $|x-a|$ then $|f(x) - L| < \epsilon$" be because someone else was sitting home drunk thinking it was funny? Well, it could have been and if it had been it wouldn't have made any difference. It's still a definition that if we can find a $\delta$ then .... we can find a delta.

But it is because we want to have a concept of a function "getting close" to a value that will be meaningful. I can't say $f(x) = x^2 + 5$ gets close to $9$ when $x$ is close to $2$ by saying "gee, just look at it here it is $x = 1.9$ that's close to $2$ and $f(x)$ is $8.61$... that's close.

So... to answer you question.

We don't care if $|f(x) -L|$ is finite. $|f(x) - L| $ is always finite. If $L = 1.7$ and $f(x)= 10^{5,198} + \sqrt {\pi}$ then $|f(x) - L| = 10^{5,198} + \sqrt{\pi} - 1.7$ which is a finite number.

We care that $|f(x)-L|$ is close to $0$. That means $f(x)$ is close to $L$. But note, we don't just want it to be close. If $f(x) = x^2 + 4$ and we want $x$ close to $2$ then we can say if $x= 2.1$ and $L = 8.97$ then $|f(x) - L| = |9.41 - 8.97| = .44$ is small. We want it to be arbitrarily small. We want to be able to get $|f(x)-L|$ to be as close to $0$ as ..... we want it to be. If we want to ght $|f(x) -L| < 0.00000000001$ and $L = 8.97$ we can't do that. There will always be an $x$ even closer to $2$ where $f(x) > 8.9700000000001$. But if $L=2$ then if we want to find to force $|f(x)-L| < 0.000000000000000000000000001$ we can. If we want to forc $|f(x) - L| < \frac 1{10^{57894375493758493758943758934}}$ we can.

So the point is... we can always get $f(x)$ arbitrarily as close as we like to $L$ by getting $x$ close to $a$. For any arbitrarily small distance between $f(x)$ and $L$ there is a small distance from $a$ that fixing $x$ within will ensure that $f(x) $ and $L$ are close.

Answer 2

One point of view thay maybe helps is to see limits as a challenge. The challenge is, given any $\varepsilon>0$, to find a $\delta>0$ that forces, when $|x-x_0|<\delta$, that $|f(x)-L|<\varepsilon$.

In other words, we say that $\lim_{x\to x_0} f(x)=L$ if, when we are given a distance from $L$, we can find a distance from $x_0$ such that for all $x$ at distance less than $\delta$ from $x_0$, we have that $f(x)$ is at less than distance $\varepsilon$ from $L$.

Answer 3

Essentially, you could phrase it as "freedom of choice of $\epsilon$ to show that $|f(x) - L| $ can be small".

It is this "freedom" that allows you to conclude that $|f(x) - L| $ can be made as small as you wish, less than any positive $\epsilon$, for a proper $\delta$.

The proof ends when we have shown the existence of $\delta$ for any choice of $\epsilon$, however small, such that $$ |f(x) - L| < \epsilon \quad\text{ for $x$ such that } \quad|x-x_0 |< \delta$$