I have a basic question about the $(\epsilon,\sigma)$ definition of limit. According to this definition, it is $\lim_{x \to c}f(x) = L$ if we have for each distance $|x-c| <\sigma$ we have an $\epsilon$ such that it is $|f(x)-L|<\epsilon$. This suggest that as $\sigma$ gets smaller and smaller we can always find a proper $\epsilon$ which limits $f(x)$'s distance from $L$. But this says nothing about the behavior of $\epsilon$ as $\sigma$ approaches to 0. One intuitively thinks that $\epsilon$ must get smaller as $\sigma$ becomes smaller but this is not said in the definition. The definition only says that we need to find an $\epsilon$. So this implies that as $\sigma$ gets smaller $\epsilon$ may get larger, which would be a weird thing to happen.

So my question is, does this definition imply $\epsilon$'s behavior as a function of $\sigma$ somehow which I fail to see? Is this possible to have a growing $\epsilon$ as the result smaller $\sigma$?

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    $\begingroup$ You have the definition the wrong way around. You specify how close you want $f(x)$ to be to $L$, then find out how close $x$ needs to be to $c$ in order to guarantee it. That is, you specify $\varepsilon$, then find a suitable $\sigma$. $\endgroup$ Commented Jul 13, 2014 at 16:51
  • $\begingroup$ While the answers below explain why the $\epsilon-\delta$ definition of limits is well-defined, I'd just like to add that I've always thought the neighborhood definition was somehow clearer (I don't really know how, since they say the same thing, but I've always thought it easier). If you know what a neighborhood is, then you could instead use this definition of limit: $\lim_{x \to a} f(x) = L \iff$ for every neighborhood $U$ of $L$ there exists a neighborhood $V$ of $a$ such that $f(V) \subset U$. If that definition makes more sense then you can use it, if not just use $\epsilon-\delta$. $\endgroup$
    – user137731
    Commented Jul 13, 2014 at 17:30
  • $\begingroup$ @Bye_World why does that definition make more sense to you? $\endgroup$ Commented Oct 23, 2016 at 1:47
  • $\begingroup$ @Pinocchio Looking back I can say that when I wrote the comment (over two years ago) I hadn't taken a class in real analysis and didn't realize how useful inequalities were. But my reasoning at the time was basically that I didn't understand the $\epsilon$-$\delta$ definition in my calculus class, but I immediately got the neighborhood definition when I saw it in multivariable calculus. Then, going back to look at the $\epsilon$-$\delta$ definition it suddenly made more sense. $\endgroup$
    – user137731
    Commented Oct 23, 2016 at 15:20

2 Answers 2

  1. The symbol that is usually used is $\delta$ (delta), not $\sigma$ (sigma), but this is not really important.

  2. I think you have the definition backwards. Here is what it should be: For any $\epsilon>0$, there exists a $\delta>0$ such that $0 < |x-c| < \delta$ implies $|f(x)-L|<\epsilon$.

  3. After making this adjustment, you are correct in thinking that $\delta$ is in some sense a "function" of $\epsilon$, and also that the definition of limit only involves existence of $\delta$, and does not consider its size. Note the following.

    • Given an $(\epsilon,\delta)$ pair that satisfies the definition of the limit, then $(\epsilon,\delta')$ would also work if $\delta' < \delta$. This shows that given an $\epsilon$, there may be many values of $\delta$ that work.
    • On the other hand, given an $(\epsilon,\delta)$ pair that satisfies the definition of the limit, then $(\epsilon',\delta)$ would also work if $\epsilon' > \epsilon$. This shows that as $\epsilon$ decreases to zero, the set of possible $\delta$ is nonincreasing. It may not be decreasing to zero (see Surb's answer), however.

If you want to compare the "rates of convergence" of $\epsilon$ and $\delta$, maybe you might want to look at the derivative? $$\lim_{x\to c} \frac{f(x)-f(c)}{x-c}$$

Edit: clarification of the second bullet in #3

Given $\epsilon>0$, let $A$ be the set of $\delta$ that satisfy the definition (explicitly, $A:=\{\delta>0:\text{$|f(x)-L|<\epsilon$, for $|x-c|<\delta$}\}$). Then if $\epsilon'>\epsilon$, then for any $\delta \in A$, we have $|x-c|<\delta$ implies $|f(x)-L|<\epsilon<\epsilon'$. So if we let $A'$ be the set of $\delta$ that satisfies the definition for $\epsilon'$ we have $A \subset A'$. Looking at this backwards, (beginning with $\epsilon'$ and then considering the smaller $\epsilon$), we see that the corresponding sets $A'$ and $A$ are nonincreasing ($A' \supset A$).

  • $\begingroup$ The two notes you give in 3 are easy to see but I did not get how they make it certain that $\delta$ is nonincreasing as $\epsilon \to 0$. How can it be shown using the properties you have listed? $\endgroup$ Commented Jul 14, 2014 at 14:03
  • $\begingroup$ To put it into a more mathematically valid statement, how can we show using the properties above for $(\epsilon_1,\delta_1)$ and $(\epsilon_2,\delta_2)$, where both satisfy the limit definition and it is $\epsilon_1 > \epsilon_2$, that it is never $\delta_2 > \delta_1$? $\endgroup$ Commented Jul 14, 2014 at 14:13
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    $\begingroup$ @UfukCanBiçici see my edits above $\endgroup$
    – angryavian
    Commented Jul 14, 2014 at 15:13
  • $\begingroup$ Thanks for the answer! So since $\epsilon' > \epsilon$ provides that for $\epsilon'$ at least all $\delta$ belonging to set $A$ satisfies the limit definition and there is always the possibility that $\epsilon'$ could extend the set $A$ with new $\delta$, so $A'$ is nonincreasing. Is that interpretation right? $\endgroup$ Commented Jul 15, 2014 at 4:24
  • $\begingroup$ @UfukCanBiçici Yes, that is right. $\endgroup$
    – angryavian
    Commented Jul 15, 2014 at 4:26

We have $$\lim_{x \to a} f(x) = L,$$ if $$\forall \epsilon > 0 \ \exists \delta > 0 \ \text{ such that } \ 0<|x-a|< \delta \text{ implies } \ |f(x)-L|< \epsilon.$$ This means that $\delta $ depends on $\epsilon$ and not the reverse. Now let $\epsilon_1 , \epsilon_2 > 0$ and the corresponding $\delta_1, \delta_2$ in the above definition. If $\epsilon_1 > \epsilon_2,$ then it is true that $$|x-a|< \max\{\delta_1,\delta_2\} \text{ implies } |f(x)-L|< \epsilon_1.$$ Now if you want an example which shows that if $\epsilon >0 $ goes to $0$ then the associated $\delta$ do not necessarily needs to go to $0$ you can consider any constant function. In this case for every $\epsilon> 0$ you may choose any $\delta > 0$. Note that in fact $\delta$ depends on $\epsilon$ and on $a$ (see uniform continuity).


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