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I am learning $\epsilon$-$\delta$ definition of limits. I was confused on a few points and read some of the related answers on this and other sites. But I couldn't find discussion on any of these questions anywhere, so I am asking them here. The questions are-

  1. Why should $|x-c|<\delta$ imply $|f(x)-L|<\epsilon$ ? Can't it be the other way round? That is, would this be wrong definition:

    " Let $f$ be a function defined on an open interval around $c$ (except possibly at $c$). Then, if for any $\epsilon>0$ there exists a $\delta > 0$ such that $0< |x-c|<\delta$ whenever $|f(x)-L|<\epsilon$; then $$\lim_{x \to c} f(x) = L$$ "?

2 Why do we choose open intervals around $x$ and $f(x)$ instead of closed ones (i.e. $x \in (x-\delta,x+\delta),f(x) \in (f(x)-\epsilon,f(x)+\epsilon)$ instead of $x \in [x-\delta,x+\delta],f(x) \in [L-\epsilon,L+\epsilon]$ ) ?

  1. Why are $\delta$ and $\epsilon$ chosen to be greater than zero? Can't we choose them to be less than zero and specify the bounds of $x$ and $f(x)$ thus: $\delta < x-c< -\delta$ and $\epsilon < f(x) - L < -\epsilon$ ?

My guess is that the answer to the points 2 & 3 should be: It is just by convention. But I have no source to back it up. I couldn't find any discussion on this anywhere. Not in my textbook, nor on Wikipedia or other sites.

So, what are the answers to these questions?

EDIT: If someone says that we can make the choices I suggested in points 2 & 3 too (instead of the ones we do currently), and so the latter are just historical conventions, please cite the source for your statement.

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  • $\begingroup$ In the first case, imagine a function that is not injective, for example, $f(x)=x^2$. We have $f(-1)=f(1)$, but $1$ and $-1$ are "far" apart (that is, they're not within $\delta$ of each other). $\endgroup$
    – Clayton
    Aug 14, 2018 at 12:31
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    $\begingroup$ For $3$, one reason to use positive quantities is so that we can simply write $|f(x)-L|<\varepsilon$ without needing a minus sign in front of $\varepsilon$. $\endgroup$
    – Clayton
    Aug 14, 2018 at 12:32
  • $\begingroup$ Question.1: $x$ approaches $c$ first, then $f(x)$ react to that by approaching $l$. That is why the definition starts with: $|x - c| < \alpha $, when that is true, then we have $ |f(x) - l | < \epsilon $. Question.2: The intervals are open because If we had $x = x - \alpha $, then $\alpha = 0$. which is not what we started with. Question.3: The definition says when $x$ is close to $c$, we have $f(x)$ is close to $l$. To express $x$ is close $c$ in the language of mathematics, we say the distance between $x$ and $c$ (which is positive) is less than $\alpha$ which needs to be positive. $\endgroup$
    – Zouhair El Yaagoubi
    Aug 14, 2018 at 12:32
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    $\begingroup$ 1. Think about $f(x) = \sin(x)$ and trying to show continuity when $L=0$. Say $c=0$. $f$ is periodic so once you find a $\delta$ such that $|f(x)-L| < \epsilon \implies |x-c| < \delta$, choose $x$ far away enough from c that still fits the first condition and you get a contradiction. $\endgroup$
    – James Yang
    Aug 14, 2018 at 12:33
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    $\begingroup$ @MrReality: Because we can say $|f(1)-1|<\varepsilon$, but $|x+1|=|x-(-1)|>\delta$ (note that in the function case, I'm using $x=1$, while in the $x$ case, I'm using $x=-1$). In essence, what we're saying is that a function might have close $y$ values where the $x$ values that map to the $y$ values are far apart. $\endgroup$
    – Clayton
    Aug 14, 2018 at 12:49

3 Answers 3

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To 2: You can do that aswell. It doesn't matter whether you work with closed or open intervals as long as you are consistent. It's equivalent.

to 3: It is because you take the absolute value of the difference which is always a non-negative. Negative $\delta$ and $\varepsilon$ wouldn't make much sense here.

To 1: Roughly speaking, taking limits mean that when you vary a little in $x$ near $c$ it should only vary a little in $f(x)$ around $L$ aswell. That's why we have an implication in here. You may draw a picture in order to understand what's happening here.

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  • $\begingroup$ Regarding your "to 3" section: Why don't they make sense? Can't we write $0<|x-c|<-\delta$ and $|f(x)-L|<-\epsilon$? $\endgroup$
    – HeWhoMustBeNamed
    Aug 14, 2018 at 12:56
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    $\begingroup$ But then, $\delta$ and $\varepsilon$ have to be negative, otherwise we have a contradiction. That means $-\delta$ and $- \varepsilon$ are positive again. But we could have chosen that from the beginning. $\endgroup$
    – YoungMath
    Aug 14, 2018 at 12:59
  • $\begingroup$ +1 for the nice and concise answer. $\endgroup$
    – Paramanand Singh ♦
    Aug 14, 2018 at 13:42
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Regarding your suggested definition of limit, I can use it to prove some nonsense, how about the following: Put $f(x) = x$. I will prove $$\lim_{x \to 1} f(x) = 2 $$

By your definition I'm asked to prove that $|x-2|<\epsilon \implies |x-1|<\delta$.

Let $\epsilon$ be given. Take $\delta = \epsilon + 1$. Then \begin{align} |&x-2|<\epsilon \\ \implies -\epsilon <& x-2 < \epsilon \\ \implies -\epsilon+1 <& x-1 < \epsilon + 1 \\ \implies -\epsilon -1 < -\epsilon+1 <& x-1 < \epsilon + 1 \qquad \text{since } -1 < 1 \\ \implies |&x-1|< \epsilon + 1 = \delta \end{align}

This is clearly not what you want from limits. You will break math this way.

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I commend a questioning mind as yours. That is how one learns mathematics.

(1) We cannot define it that way because it would mean another thing. It would mean that no matter how close $f(x)$ is to $L,$ we can find a single $\delta>0$ so that $x$ remains fixed to within $\delta$ of $c.$ Clearly, $x$ is not approaching $c$ or anything here, so that that suggestive arrow means nothing here, or is at best misleading. Furthermore, this definition does not capture what we want to mean by the limit of a function as the independent variable approaches a fixed point $c.$ The idea is that given any $\epsilon>0$ we can find a $\delta>0$ (for each given $\epsilon>0,$ i.e., this $\delta$ need not be the same for all the $\epsilon$'s) so that whenever $x$ is within $\delta$ of $c,$ we must have $f(x)$ to be within the pre-assigned $\epsilon$ of $L.$ Basically, we want the behaviour of $f(x)$ near $L$ to be controlled by the corresponding behaviour of $x$ near $c.$

(2) First, I should mention that it is around $c$ and $L$ that we require open intervals wherein $x$ and $f(x)$ range respectively. Well, one reason is that (just like we do not require that $f$ have a value at $c$) we also do not want to think of boundaries. That is, we want to emphasise that we are only concerned with how $f$ behaves near $c$ -- not at $c$ or at the boundaries of the interval. So even if the function is defined on the closure of the interval, we prefer to use any open interval about $c$ (indeed, it's usually thought of as a very small open interval to emphasize the locality of this idea).

(3) We do not choose negative $\epsilon$ and $\delta$ because we think of them as distances, and we do not want our distances to be negative (in higher analysis, we formalise these ideas by defining a metric space as a set together with a distance function; one of the requirements on this function is that it be positive-definite; that is, all distances are positive except when the points coincide, when the distance vanishes).

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