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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?

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    @thei: Agreed. Though question is really about the typical $\epsilon-\delta$ definition as given in most texts (I've seen).2011-09-17

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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.

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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.2011-09-17
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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.

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    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.2011-09-15
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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.

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    @raxa: yes. (In case of further discussion let us do it in comments under the question so as not to constantly ping Robert.)2011-09-16