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?

  • $\begingroup$ Edited to correct an overstatement of (1). $\endgroup$ – orome Sep 15 '11 at 19:01
  • $\begingroup$ Would this be clearer if I replaced my $>0$s with $≠0$s? $\endgroup$ – orome Sep 15 '11 at 19:04
  • $\begingroup$ No, $\gt$ is correct. The absolute values are always positive, so $\delta$ and $\epsilon$ better be, too. $\endgroup$ – Ross Millikan Sep 15 '11 at 19:53
  • $\begingroup$ I've restated the question to emphasize the parallel structure of the implications of (1) and (2) — as I understand them. $\endgroup$ – orome Sep 15 '11 at 19:57
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    $\begingroup$ @rax: I posted a sort of meta-analysis. $\endgroup$ – zyx Sep 16 '11 at 23:23

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.

  • $\begingroup$ 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. $\endgroup$ – orome Sep 17 '11 at 17:01

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.

  • $\begingroup$ Actually for (2.), all functions will fail, except those that are identically equal to $L$ in some neighborhood of $a$. :-) $\endgroup$ – Srivatsan Sep 15 '11 at 18:31
  • $\begingroup$ @Srivatsan Narayanan: Even those will, given the less than sign. No absolute value is $\lt 0$ $\endgroup$ – Ross Millikan Sep 15 '11 at 18:35
  • $\begingroup$ Oh, you are right. Sorry about that. $\endgroup$ – Srivatsan Sep 15 '11 at 18:36
  • $\begingroup$ @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! $\endgroup$ – orome Sep 15 '11 at 20:04
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Sep 15 '11 at 20:39

For (3), if $\delta = 0$ was allowed the definition would apply to everything: since $|x-a| < 0$ is impossible, it implies whatever you like.

  • $\begingroup$ 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? $\endgroup$ – orome Sep 15 '11 at 21:45
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    $\begingroup$ 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 $\delta = 0$). $\endgroup$ – zyx Sep 15 '11 at 23:34
  • $\begingroup$ @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? $\endgroup$ – orome Sep 15 '11 at 23:40
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    $\begingroup$ 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. $\endgroup$ – zyx Sep 15 '11 at 23:47
  • $\begingroup$ @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?) $\endgroup$ – orome Sep 16 '11 at 0:52

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