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What are the consequences of the three nonzero requriments in the definition of the limit:

$\lim_{x \to a} f(x) = L \Leftrightarrow \forall$ $\varepsilon>0$, $\exists$ $\delta>0 :\forall$ $x$, $0 < \lvert x-a\rvert <\delta \implies \lvert f(x)-L \rvert < \varepsilon$

I believe I understand that:

  1. if $0 = \lvert x-a\rvert$ were allowed the definition would require that $f(x) \approx L$ at $a$ ($\lvert f(a)-L \rvert < \varepsilon$);

  2. if $\varepsilon=0$ and $\lvert f(a)-L \rvert \le \varepsilon$ were allowed the theorem would require that $f(x) = L$ near $a$ (for $0 < \lvert x-a\rvert <\delta$); and

  3. if $\delta=0$ were allowed (and eliminating the tautology by allowing $0 \le \lvert x-a\rvert \le \delta$) the definition would simply apply to any function where $f(a) = L$, regardless of what happened in the neighborhood of $f(a)$.

Of course if (2'.) $\varepsilon=0$ were allowed on its own, the theorem would never apply ($\lvert f(a)-L \rvert \nless 0$).

What I'm not clear about is [A] the logical consequences of (3'.) allowing $\delta=0$ its own, so that:

$\lim_{x \to a} f(x) = L \Leftrightarrow \forall$ $\varepsilon>0$, $\exists$ $\delta≥0 :\forall$ $x$, $0 < \lvert x-a\rvert <\delta \implies \lvert f(x)-L \rvert < \varepsilon$

and [B] whether allowing both 1. and 2. would be equivalent to requiring continuity?

William
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orome
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  • Edited to correct an overstatement of (1). – orome Sep 15 '11 at 19:01
  • Would this be clearer if I replaced my $>0$s with $≠0$s? – orome Sep 15 '11 at 19:04
  • No, $\gt$ is correct. The absolute values are always positive, so $\delta$ and $\epsilon$ better be, too. – Ross Millikan Sep 15 '11 at 19:53
  • I've restated the question to emphasize the parallel structure of the implications of (1) and (2) — as I understand them. – orome Sep 15 '11 at 19:57
  • I've restated again to focus on the questions, A and B. The first of with is really about the logical structure of the theorem (I may be getting over my head). – orome Sep 15 '11 at 21:47
  • @zyx: Continuing discussion below per your suggestion. Putting it that way (my comment below) also highlights (for me) the important additional consequences tinkering with the definition of limit following (3'): not only would such a "limit" always exist for every function, it would exist everywhere and could be anything you want it to be. – orome Sep 16 '11 at 21:36
  • @zyx:
    I'd like to update and accept the answer Robert gave (which I is why I was commenting on his answer) to look something like the last couple of comments I made. I'd then mark that as the accepted answer. How do I go about that, considering that at least you and Robert are contributors?     – orome Sep 16 '11 at 21:53
  • @raxa: I'm not sure what the etiquette protocols are for this website, but I guess accepting Robert's answer, then posting any updates as your own answer (which could then absorb any followup discussion in the comments below it). This avoids editing Robert's text though again I am not sure whether the site encourages or discourages users editing each others' answers. – zyx Sep 16 '11 at 22:09
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    @rax: I posted a sort of meta-analysis. – zyx Sep 16 '11 at 23:23
  • @zyx: Looks great! (Though I'd never have understood it without the discussion I've had here, I now do.) – orome Sep 16 '11 at 23:28
  • The non-zero requirement on the distance between x and a is actually quite a bad way to state this condition. The good way is to have equality and to write "for all x where f is defined". – Phira Sep 16 '11 at 23:34
  • @thei: more is needed or log(x) would have limits for x < 0. – zyx Sep 16 '11 at 23:45
  • @zyx Well, yes. This corresponds to the fact that you can add arbitrary isolated values for negative numbers to $log(x)$ without destroying continuity. – Phira Sep 17 '11 at 08:48
  • @thei: Agreed. Though question is really about the typical $\epsilon-\delta$ definition as given in most texts (I've seen). – orome Sep 17 '11 at 16:10

3 Answers3

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For (1) we don't even want to require that $f$ be defined at $x=a$. Think of $\lim_{x=0}\frac{x^2}{x}$, which we would like to have limit $0$. For (2) if we allow $\epsilon$ to be $0$ then the absolute value would always fail. Your idea about (3) is spot on.

Ross Millikan
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  • Actually for (2.), all functions will fail, except those that are identically equal to $L$ in some neighborhood of $a$. :-) – Srivatsan Sep 15 '11 at 18:31
  • @Srivatsan Narayanan: Even those will, given the less than sign. No absolute value is $\lt 0$ – Ross Millikan Sep 15 '11 at 18:35
  • Oh, you are right. Sorry about that. – Srivatsan Sep 15 '11 at 18:36
  • @Ross: Can you have a look at the reformulated question? My initial formulation may have confused what I was asking with regard to (2). Thanks! – orome Sep 15 '11 at 20:04
  • I think your original versions of (1) and (2) were better. As I said, if $x=a$ is allowed the function must be defined there (and it would have to be close to $L$), which we don't want to require. For (2) as $|f(x)-L|$ must be strictly less than $\epsilon$ no function would be continuous if $\epsilon =0$ were allowed. – Ross Millikan Sep 15 '11 at 20:39
  • @Ross: OK, I've incorporated the cases where the theorem just wouldn't apply into the question, to focus on my original intent, which was about the consequences of relaxing these requirements, not (as I originally asked) the reasons for having them. My original formulation did not make that clear (and had some mistakes). – orome Sep 15 '11 at 21:02
  • For reference interpreting the above comments: the (2) referred to there is now (2'). – orome Sep 15 '11 at 22:12
  • I think your statements of 2 and 3 are now correct and in the right spirit. For 3', allowing $\delta=0$ after the there exists changes nothing because the $\lt$ signs in the later clause force $\delta \gt 0$. Think about whether there is any difference if you put $\exists \delta \gt -1$ there. Nothing happens because it won't satisfy the following inequality. – Ross Millikan Sep 15 '11 at 22:32
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For (3), if $\delta = 0$ was allowed the definition would apply to everything: since $|x-a| < 0$ is impossible, it implies whatever you like.

Robert Israel
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  • This is the crux of my question (A), which is really about the logical structure of a theorem and the way it is applied, so let me make sure I understand. Take a theorem that says that something, $D$, is defined when some values $v$ exist such that $P(v) \implies Q$. What happens when $P(v)$ is impossible for some allowed $v$. Does the theorem then apply to everything? – orome Sep 15 '11 at 21:45
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    The definition requires that any $x$ satisfying one property (dependent only on $x$) satisfies another (depending on both $x$ and the function). Each $x$ satisfying the first property can be considered as a test to which the function is subjected: does the function give this $x$ the required second property. If the set of tests that the function must pass is empty, it passes automatically. That is what happens when no $x$ exists satisfying the first property ($|x-a|<\delta$, when $\delta = 0$). – zyx Sep 15 '11 at 23:34
  • @zyx: That's what I'm looking for. Should it be obvious to me that when the set of tests that must be passed is empty, they are passed automatically? – orome Sep 15 '11 at 23:40
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    I don't know if it's obvious, any more so than other conventions from sentential logic, but it is a convention that allows boundary cases to be covered smoothly without explicit exception handling. In this case it is formally equivalent to the principle that a product of elements in an empty set of numbers is equal to 1 (as in $0! = 1$) or the principle that an empty sum is zero. – zyx Sep 15 '11 at 23:47
  • @zyx: But if a thing exists iif something else exists that satisfies a condition, and that condition can never be satisfied, it doesn't make sense to conclude that the thing exists. Does it? (And in any case, doesn't this only matter of we require $\delta = 0$, and not $\delta \le 0$ (as in 3') which, as Ross says, would be a no-op?) – orome Sep 16 '11 at 00:52
  • @raxa: here a thing $\delta$ satisfying a condition $C(\epsilon,f,L,a)$ exists if and only if something else (a nonempty set or sequence S of values of $x$) that satisfies another condition $D(f,L,a)$ does not exist. In the case where $\delta=0$ is allowed, condition D can never be met so that $\forall \epsilon \exists \delta \dots$ is satisfied automatically by taking $\delta = 0$ for each $\epsilon$. – zyx Sep 16 '11 at 05:07
  • @zyx: It's me then (or my newness to the subject), but to me it looks like $D$ needs to exist. Could you spell out precisely what $D$ is for me? Thanks! – orome Sep 16 '11 at 12:37
  • @zyx: Aha! I've got it now. Proving (no pun) my ignorance, I had missed the simple fact that $F \implies q$ is always $T$. If that's the case, then the answer to [A] is clear: allowing $\delta = 0$ means we can choose $\delta = 0$ resulting in $0<0 \implies q$, which is always true which means that the definition applies to all $f$. Do I have that right (at last)? – orome Sep 16 '11 at 18:55
  • Yes, but be aware that scope of the F-->T thing is a statement about values of $x$, not $f$. The consequences for $f$ ("the definition applies to all $f$") come about by unwinding the implications of the F-->T back through the other quantifiers, from $\delta$, to $\epsilon$, and finally noticing that the end result would be independent of $f$ and $L$. Also, for the usual definition with $\delta > 0$, the object "S" that has to not exist is a sequence of values $x_i$ converging to $a$, at which $|f(x_i) - L| \geq \epsilon$. Consider $f=\sin(1/x), a=L=0$ and $\epsilon=100$ vs $\epsilon=1/2$. – zyx Sep 16 '11 at 19:54
  • @zyx: Making sure I have the unwinding right: if $\delta=0$, $(0 < \lvert x-a\rvert <\delta \implies \lvert f(x)-L \rvert < \epsilon$) is satisfied $\forall x$, regardless of $a$, $f$, $L$, and $\epsilon$; and is therfore true $\forall \epsilon$; so $\lim_{x \to a} f(x) = L$ always (i.e., for any $a$, $f$, $L$). – orome Sep 16 '11 at 21:12
  • @raxa: yes. (In case of further discussion let us do it in comments under the question so as not to constantly ping Robert.) – zyx Sep 16 '11 at 21:24
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The question has been answered, but for sorting out the $(2^5 - 1)$ different ways of replacing strict inequalities by weak ones in the definition, the following might help.

The condition to be met is more stringent for smaller $\epsilon$. If you allow $\epsilon \geq 0$ there is no need for the $\forall \epsilon > 0$ quantifier, one can just replace $\epsilon$ by $0$ everywhere in the definition. The logical formula will then either express the condition that a function be equal to $L$ on a neighborhood of $a$, or be so strict that no function meets the condition. Assume, then, that the formula begins $\forall \epsilon > 0 \dots \quad$. In that case it makes no difference whether in the final inequality $|f(x)-L|$ is $ < \epsilon$ or $\leq \epsilon$.

The condition to be met is less stringent for smaller $\delta$. If $\delta =0$ is allowed then the $\exists \delta \dots$ can be satisfied if and only it is satisfied by $\delta=0$, and one can replace $\delta$ by zero everywhere instead of quantifying over $\delta$. In that case one gets either a condition that is true for every function, or the condition that $f(a)=L$, according to whether $x=a$ is allowed.

The requirement that $0 < |x-a| < \delta$ is the one that is most natural to modify. It defines the type of neighborhood of $a$ on which the convergence to $L$ occurs. Here it is a punctured two-sided neighborhood (usually to allow discussion of derivatives where ratios of type 0/0 appear, like $\sin(x)/x$ near $x=a=0$) but allowing $x=a$ gives a definition of continuity, or one might want one-sided limits with $ 0 < x-a < \delta$ or $0 < a - x < \delta$. If $\delta=0$ is permitted then the natural neighborhood to use would be $0 \leq |x-a| \leq \delta$ but this would only lead to a complicated restatement of "$f(a)=L$". Finally, changing the upper bound to $|x-a| \leq \delta$ would not affect anything (except in the useless case where $\delta=0$ is allowed).

To summarize, allowing $0 \leq |x-a|$ gives a definition of continuity, but changes to any of the other inequalities $\epsilon > 0$, $\delta > 0$, $|x-a| < \delta$ or $|f(x)-L| < \epsilon$ either do not affect the definition, or trivialize it.

zyx
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  • As the question and the discussion has evolved, this is now the most suitable answer, and I've marked it as such. Could you remove the "question has been answered" part and, if it's not too much trouble insert references to the numbers and letters in the question? I'd do it but I don't have edit privileges? Thanks. – orome Sep 17 '11 at 17:01