I Explicit logical justification for last step in epsilon/delta proof?

[nomadreid]

TL;DR Summary

How the assumptions allow one to choose delta between (a chosen numerical arbitrary upper bound) and (an upper bound derived from that numerical upper bound) in a standard epsilon-delta limit proof is unclear to me.

In the last step in a standard technique for doing an epsilon-delta proof of a limit (in which one takes the minimum of two possible functions delta of epsilon) the presentations I have found usually have a logical jump that I am not sure how to fill in, despite staring at examples.

The general idea for the limit of the function as x approaches a equals L is usually to

(a) factor the |f(x)-L| into |x-a|*|h(x)| < epsilon
(b) select an arbitrary positive k so that |x-a|< delta < k
(c) thus find an upper bound b so that |h(x)|< b
(d} thus b*delta < epsilon, or delta < epsilon/b
(e) for delta, pick the minimum of k or epsilon/b

Then one shows that each of the choices works by plugging them into the definition. Fine. Nonetheless, explicitly justifying [e] is always lacking, presumably because it is obvious. Alas, "obvious" is relative. To put it another way: how would one break up the second "implies" in either of the following formulations?

(|x-a|< k implies that delta < epsilon/b) implies (delta < k or delta < epsilon/b).
Alternatively,
(|x-a|< delta implies that delta < epsilon/b) implies (delta < k or delta < epsilon/b).
(If I got some details in my exposition wrong, corrections would be super nice, as well as to answer the question with its errors corrected.)

Many thanks for those with patience, as I am sure this question has been asked before (although I did not find it).

 

[mjc123]

For future reference, the letter B in square brackets is code for bold text. If you want a B in square brackets in your text, include a space, as [B ]. (Sorry, I don't know the answer to your question.)

 

[nomadreid]

Thanks for that, mjc123. (Oops.) I have edited it.

 

[pasmith]

If |x - a| &lt; \delta &lt; k is sufficient, then we are done; if it is not, then we must be looking for 0 &lt; \delta &lt; k, and we can assume that in our calculations. Having thus concluded that in this case 0 &lt; \delta &lt; b/epsilon &lt; k, we can put these two cases together to find that \delta &lt; \min\{k, b/\epsilon\}.

 

[nomadreid]

Thanks very much, pasmith.

First, I apologize for writing out delta, etc., but I have attempted to type it with \delta as seems to be indicated in the "LaTeX Guide" on this forum, but without the required results. I also tried ##\delta and \delta\ without success, and even \delta. Sorry.<br /> <br /> I believe I understand the details; my confusion apparently came from expositions which used &quot;delta&quot; in the same set of equations for more than one meaning.<br /> <br /> Here is how I understand your explanation if I pedantically fill in some of the details (using the same notation as above)<br /> <br /> For a given epsilon:<br /> <br /> [1] if |x-a|&lt;k implies |f(x)-L| &lt; epsilon, then delta = k suffices<br /> <br /> [2] If not: if |x-a|&lt;k, then h(x)&lt;b.<br /> <br /> Then if |x-a|&lt;epsilon/b and h(x)&lt;b then |x-a|*h(x)= |f(x)-L| &lt;epsilon, so set delta= epsilon/b<br /> <br /> [3] If |x-a|<u>&gt;</u>k, then there is an m&gt;k such that |x-a|&lt;m; repeat [2] and we would get delta=epsilon/c<br /> <br /> for some c&lt;b, so that epsilon/c &gt; epsilon/b, so in this case epsilon/b would still work.

 

[fresh_42]

I don't want to interfere with the tutorial here. But in case you are interested in a visual presentation and on my take on what continuity is, then you can find a picture and some explanations here:
https://www.physicsforums.com/insights/epsilontic-limits-and-continuity/

 

[nomadreid]

fresh_42, thanks a lot! This looks like it will be very useful.

 

[WWGD]

Well, i hope I understood your question well, but it's justified by the fact that $\delta$is a function of $\epsilon$. If f Is differentiable, a good rule of thumb is used by $f'(x)$ for the relation between the two. Is that what you were asking?

 

[nomadreid]

Thanks for the reply, WWGD, but my question is a bit more basic.

In a nutshell, I am asking whether the process [1],[2],[3] in post #5 is a proper explicit explanation of why the last step in the typical epsilon-delta proof of a limit is valid: that is, when you assume a bound for delta at the beginning and end up with the minimum between two quantities as the last step. The usual explanations I have found seem rather hand-wavy to me; I am trying to get it explicit.

This is not directly about derivatives. (Of course, since derivatives are defined in terms of limits, then of course it is also indirectly about them.)

 

[Gavran]

(a) factor the |f(x)-L| into |x-a|*|h(x)| < epsilon
(b) select an arbitrary positive k so that |x-a|< delta < k
(c) thus find an upper bound b so that |h(x)|< b
(d} thus b*delta < epsilon, or delta < epsilon/b
(e) for delta, pick the minimum of k or epsilon/b

Your five steps are correct except the second one which should be:
(b) select an arbitrary positive k so that |x-a|< k.

It is not always possible to find b where |h(x)|<b because the function h(x) can be unbounded. This problem can be solved by assuming that x is bounded, so there is |x-a|< k. In other words, you choose k in the way that h(x) is bounded for |x-a|< k.

At the end there are two conditions |x-a|< k and |x-a|<epsilon/b and both of them must be satisfied, so there is delta=min{k,epsilon/b}.

 

[nomadreid]

Thanks very much for the correction and explanation, Gavran.

Your correction leaves me a little puzzled, in that all the expositions and examples I have read derive the relationship between epsilon and delta by assuming that
|x-a|<delta.

That is, they
factor the |f(x)-L| =h(x)*|x-a| <epsilon,
get a limit for h(x) based on the chosen limit for delta
and then use both bounds to get
b*delta as in step 4.
With your correction disallowing an assumed bound for delta, I no longer see how that is possible.

A sample of what I mean -- so that perhaps it will be clearer as to what I am misreading:
the first answer in
https://math.stackexchange.com/questions/286078/proving-a-limit-using-epsilon-delta-definition

 

[pasmith]

The implication is (|x - a| &lt; \delta &lt; B) \implies (|f(x) - f(a)| &lt; K\delta). Thus to be sure that |f(x) - f(a)| &lt; \epsilon we need both \delta &lt; B and \delta &lt; \epsilon/K, ie. \delta &lt; \min(B,\epsilon/K). That's really all there is to it.

 

[nomadreid]

Thanks, pasmith.

I am wondering about the apparent difference between your explanation and that of Gavran (post #10) who told me (if I understood correctly) to eliminate the delta in the premise of the inequality in your implication.

 

[Gavran]

Thanks, pasmith.

I am wondering about the apparent difference between your explanation and that of Gavran (post #10) who told me (if I understood correctly) to eliminate the delta in the premise of the inequality in your implication.

I have followed the next $$ |f(x)-L|=|x-a|\cdot|\frac{f(x)-L}{x-a}|\lt\delta\cdot b=\epsilon $$ where $$ |\frac{f(x)-L}{x-a}|\lt b $$ for $|x-a|\lt k$.

$\delta=\min\{k,\frac\epsilon b\}$

This is not the general approach. There are other approaches which also use $\delta$ and $\epsilon$ to prove that $$ \lim_{x\to a}f(x)=L $$.

 

[nomadreid]

Thanks, Gavran. If I get this straight, to set b you can use the bound k for |x-a|, allowing you to set delta in terms of epsilon and b. But then where does the bound delta<k come from in the last step, and what role does it play?

 

[Gavran]

Thanks, Gavran. If I get this straight, to set b you can use the bound k for |x-a|, allowing you to set delta in terms of epsilon and b. But then where does the bound delta<k come from in the last step, and what role does it play?

You can guarantee that $|f(x)-L|\lt\epsilon$ only for two cases: $|x-a|\lt k\le\epsilon/b$ and $|x-a|\lt\epsilon/b\le k$.
Can you guarantee that $|f(x)-L|\lt\epsilon$ for $k\lt|x-a|\lt\epsilon/b$?

 

[PeroK]

Thanks very much for the correction and explanation, Gavran.

Your correction leaves me a little puzzled, in that all the expositions and examples I have read derive the relationship between epsilon and delta by assuming that
|x-a|<delta.

That is, they
factor the |f(x)-L| =h(x)*|x-a| <epsilon,
get a limit for h(x) based on the chosen limit for delta
and then use both bounds to get
b*delta as in step 4.
With your correction disallowing an assumed bound for delta, I no longer see how that is possible.

A sample of what I mean -- so that perhaps it will be clearer as to what I am misreading:
the first answer in
https://math.stackexchange.com/questions/286078/proving-a-limit-using-epsilon-delta-definition

Let me show you how I would think about that example. We want to show that:
$$\lim_{x \to 2} (x^2 + 2x - 7) = 1$$What we are interested is the expression $|(x^2 + 2x -7) - 1|$, which as we can all calculate is equal to $|x + 4||x-2|$.

When $x$ is close to $2$, $|x + 4|$ is close to $6$. Informally, as $x \to 2$, we have:
$$|(x^2 + 2x -7) - 1| \approx 6|x- 2|$$So, informally, for small $\epsilon$, we need to choose $\delta \approx \frac \epsilon 6$.

We need to formalise this. There are two ways to do this.

a) The first method is to note that we don't have to worry about large $\epsilon$. In general, the following is equivalent to the normal definition of a limit:
$$\forall 0 < \epsilon < 1, \ \exists \delta > 0: \ 0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$$And, in fact, that holds for any other number than $1$, as the maximum $\epsilon$ we need to concern ourselves with. So, in general, you can always assume $\epsilon < 1$.

b) The second method is to note that we don't have to worry about large $\delta$. Again, we can modify the standard definition to the equivalent:
$$\forall 0 < \epsilon, \ \exists 1> \delta > 0: \ 0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$$Again, in general we can always assume $\delta < 1$. Why would we ever consider $\delta > 1$? Why would we ever worry about choosing a large $\delta$?

In this case, let's use the second approach by assuming that $\delta < 1$, we know that $1 < x < 3$ and $5 < |x + 4| < 7$. And, we can see that what we want is $\delta = min\{1, \frac \epsilon 7\}$

 

[nomadreid]

Thanks very much, Gavran. As far as I understand, you are saying that one derives a delta based on both cases, and then pick the smallest of the two? I will work through some cases this way; it is an interesting approach that was not explicitly presented in any of the examples which I have seen.

 

[PeroK]

PS I read through the stack exchange page. And, in fact, the last poster took an approach similar to mine. What I don't like about the way epsilon-delta is generally taught, is that you are asked to consider arbitrary positive epsilon and delta. Whereas, what you are interested in is arbitrarily small epsilon and small delta.

A lot of time and effort seems to be wasted worrying about the cases where epslion and/or delta is large. Those cases are never an issue for a limit. You can always put $1$ as an upper bound on them both whenever you need to. Or, choose any other fixed positive upper bound.

 

[PeroK]

PPS note that if we choose the first method and explicitly consider only $\epsilon < 1$. Then, after the preliminary calculations, we can simply choose $\delta = \frac \epsilon 7$, as in this case we automatically have $\delta < 1$.

 

[nomadreid]

Thanks very much, PeroK. That is very helpful The idea of only considering small delta and epsilon is completely reasonable. (It makes me think that the proofs could be adapted to a non-standard analysis approach in requiring that delta and epsilon are infinitesimals.)