WHY are we always given $\epsilon > 0$ first, then solving for a $\delta>0$? This is in the limit definition.

I want to ask:

Can we say "given $\delta>0$, there exists $\epsilon>0$"? Since we can always solve for one given the other.

I found three counterexamples, but I don't understand them:

  1. Let $f(x) = \sin x$, let $L$ and $\delta$ be arbitrary real numbers. Then $\epsilon = |L| + 2$ satisfies your definition. (from post)

    Q: What's wrong with setting $\epsilon = |L| + 2$? It's big, but it's not wrong!

  2. Let $f(x) = 1/x$, and let $a = 1$. The definition fails for $\delta \ge 1$, since for any $\epsilon$ we can choose $x=1/(L+\epsilon)$ if $L+\epsilon > 1$, so that $f(x)-L \ge \epsilon$. (from post)

    Q: What are they saying here? At $x=1$, the definition fails for $\epsilon \ge 1$ too! The problem is not $\delta$. The problem is the function is undefined for $x \le 0$.

  3. Counterexample: $\lim\limits_{x \to 0} f(x) = L$

    $f(x) = \begin{cases} \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$

    Given any $\delta > 0$, we can find $\epsilon > 0$ such that $|f(x) - L| < \epsilon$ whenever $|x| < \delta$. For instance, set $\epsilon = 2$; then any choice of $L \in (-1,1)$ will satisfy this "reversed" situation. (from post)

    Q: I don't see how setting $\epsilon = 2$ violates any definition. I mean, we did find a $\epsilon$ for a given $\delta$.

Thanks all for the pouring answers, I'll get back to each one personally. If I did not choose an answer, that means all submissions are still welcomed! The best answer will be chosen based on # of upvotes (50%) and if I understood it and agree it's the best (50%).

  • 1
    $\begingroup$ Epsilon and delta are in not one to one correspondence. Consider step function ( greatest integer function) and assume for any delta epsilon is 100. Now see what happens $\endgroup$
    – Cloud JR K
    Feb 20, 2020 at 22:02
  • 6
    $\begingroup$ In a notational sense, you can use $\delta$ in place of $\varepsilon$ and vice-versa, but you cannot reverse the roles that these values play (and you need to use the right quantifiers in the right order). $\endgroup$
    – Dave
    Feb 20, 2020 at 22:03
  • $\begingroup$ As @Dave mention if it's about notation then its doesn't really matter but if not (then it's a disaster if we change it $\endgroup$
    – Cloud JR K
    Feb 20, 2020 at 22:06
  • 1
    $\begingroup$ The whole point of limit "$\lim_{x\to a}f(x)=L$" is $f(x)$ become arbitrarily close to $L$ when $x$ is sufficiently close to $a.$ Careful thought of this idea would convince you that $\epsilon$ is arbitrary, but $\delta$ is not. And there is no $1-1$ correspondence between them. $\endgroup$
    – Bumblebee
    Feb 20, 2020 at 22:10
  • 1
    $\begingroup$ @RayLittleRock That's true. If we picked $\delta$ first, we can always pick a bigger $\epsilon$, which wouldn't make the bound tight. I answered here: math.stackexchange.com/a/3560152/44802 $\endgroup$
    – user13985
    Mar 11, 2020 at 16:22

8 Answers 8


The point of the definition of limit is to capture the idea that we can force the values of $f$ to be close enough to $L$, provided only that the values of $x$ be sufficiently near $a$: if you tell me how close you want $f(x)$ to be to $L$, I can guarantee this outcome by telling you how close $x$ should be to $a$.

Reversing the logical dependency between $\epsilon$ and $\delta$ makes the logical dependency run the "wrong" way: you are saying that you will tell me how close you want $x$ to be to $a$, and then I will be forced to tell you how close I can guarantee $f(x)$ to be to $L$.

It might seem like this would be good enough, but it doesn't work. It seems to be the same because you may be thinking of the limit as saying "the closer you get to $a$, the closer the values will get to $L$"; but saying the limit is $L$ is more than that: it says that values get arbitrarily close to $L$, and that all values get close to $L$ near $a$, not just some.

If you are allowed to pick the value of $\epsilon$, then you are not guaranteeing that the values get arbitrarily close to $L$, just that they get "sufficiently" close to $L$.

So for example, you want the limit of $f(x)$ to approach at most one thing, not two or more. But say that $f(x)$ always takes values between $-1$ and $1$, as $f(x)=\sin(x)$ does. If I take $L$ to be any value between $-1$ and $1$, and then let $\epsilon=3$, then regardless of what your $\delta$ is, we will indeed satisfy that $|f(x)-L|\lt \epsilon$ whenever $|x-a|<\delta$. So every number between $-1$ and $1$ is a limit. And worse, any number is a limit: if you give me $L=10$, then provided I let $\epsilon>11$, every value of $f(x)$ will be within $\epsilon$ of $L$.

That means this definition doesn't really capture the notion we want the definition of limit to capture.

Remember: to convince me that the limit is $L$, you challenge me to fall within an arbitrarily thin horizontal band around $L$. The challenge lies in how thin the horizontal band it; if you let me pick how thin that band is, then I can make it really fat and have absolutely no challenge at all.

  • $\begingroup$ This is bravo! This right here read my mind: "you may be thinking of the limit as saying 'the closer you get to 𝑎, the closer the values will get to 𝐿.'" But still not totally convinced yet. $\endgroup$
    – user13985
    Feb 21, 2020 at 3:47
  • 2
    $\begingroup$ @user13985: “How do you know will it get closer?” Sorry, but there’s too many quantities about. You are going to have to specify what “it” and closer to what you are talking about. As to your objection, not that it says that for any $\epsilon$, no matter what it is, you can find a $\delta$. No matter how narrow you make the horizontal band around $L$, there is a way of selecting a vertical band around $a$ so that all values inside the vertical band lie in the horizontal band. That’s how I “know” the values are getting closer to $L$: because I can guarantee they are as close as I want. $\endgroup$ Feb 21, 2020 at 4:29
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    $\begingroup$ @user13985: Finally, remember that the $\epsilon$-$\delta$ definition tries to recapture the infinitesimal notions (because infinitesimals don’t exist in the usual real numbers): the value of $f(x)$ will be “infinitesimally close” to $L$, provided that $x$ is “infinitesimally close” to $a$. You tell me how close you want the value of $f$ to be to $L$, I tell you how close $x$ has to be to $a$ to make sure the value is that close. $\endgroup$ Feb 21, 2020 at 4:31
  • 1
    $\begingroup$ @user13985: The $\delta$ can't both be too large and successful. Once the $\epsilon$ gets small enough, I will be forced to pick $\delta$. I may pick it smaller than needed, but I will be forced to pick to small if you give me a small $\epsilon$. In your challenge-response, I don't even need to know what your requirement is, $\epsilon$ can be picked arbitrarily and large enough to fit everything. And if you can pick $\epsilon$ to fit everything, then you aren't really getting any information about the function at all. $\endgroup$ Feb 21, 2020 at 15:22
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    $\begingroup$ @user13985: In the usual definition, I can prove that the limit of $\sin(x)$ as $x\to 0$ is $0$ by showing that no matter what $\epsilon$ you pick, I can find an $\delta$ that will work. In your definition, I can likewise prove that the limit is 17 by showing that no matter what $\delta$ you pick, I can find an $\epsilon$ that will work. In the usual definition, I can prove the limit is not $17$ by showing that no matter what $\delta$ you pick, you can never make it work for $\epsilon=1$. In your definition, you *cannot prove the limit is not $17$, or $1000$, or $-3$. $\endgroup$ Feb 25, 2020 at 4:01

The definition of a limit involves the sentences of the form "for every $\varepsilon>0$ there exists $\delta>0$ such that...". It is not possible to replace this with "for every $\delta>0$ there exists $\varepsilon>0$ such that..." - you get a different, and certainly not equivalent sentence.

This is a purely logical argument. "for every $x$ there exists $y$" is not the same as "for every $y$ there exists $x$". You can find examples from everyday life:

  • "every book has been written by a human" is not the same as "every human has written a book"
  • "every monkey lives in a jungle" is not the same as "in every jungle there lives a monkey"

and so on and so forth.

For a more mathematical example: "for every real number $x$ there is a real number $y$ such that $y=x^2$" (which is true) is not the same as "for every real number $y$ there is a real number $x$ such that $y=x^2$" (which is false).

  • $\begingroup$ That makes some sense. But there is a mapping $f: x \rightarrow y$. In particular, you had $y=x^2$. And y is not uniquely mapped to a single x, which is fine. Because the $\delta$-$\epsilon$ definition just says there has to exists just one $\delta$. $\endgroup$
    – user13985
    Feb 20, 2020 at 23:05
  • $\begingroup$ @user13985 : When you say $\delta-\epsilon$ in your previous comment - do you mean "standard" definition or do you mean your definition with swapped $\epsilon$ and $\delta$? If you mean the standard definition, as it seems to me, then no - the standard definition does not imply uniqueness of $\delta$ - it just needs to exist. There may be many $\delta$'s for any given $\epsilon$ (and usually, when you find such $\delta$, any smaller $\delta$ would also do). $\endgroup$
    – user700480
    Feb 20, 2020 at 23:13
  • $\begingroup$ I meant the standard/regular definition. So, you shouldn't require my swapped definition to be unique either. In your example $y=x^2$, it seems to suggest my swapped definition failed, because there's no unique epsilon for my delta. $\endgroup$
    – user13985
    Feb 20, 2020 at 23:17
  • $\begingroup$ @user13985 Not just that there isn't a unique one - there may not be any one. Set $y=-1$. $\endgroup$
    – user700480
    Feb 20, 2020 at 23:18
  • $\begingroup$ You answer emphasized the difference between injections and surjections. $\endgroup$
    – user13985
    Feb 20, 2020 at 23:19

Well, you can say given a $\delta$ there exists an $\epsilon$. It's just that you DO anything interesting with that result.

Let's suppose we want to prove that $\lim_{x\to \frac \pi 4} \sin x = 5$.

That is obviously not true. So if we can prove it we have obviously done something wrong.

Well let's let $\delta$ be any $\delta> 0$. And let's let $\epsilon > 7$.

Whenever $|x- \frac \pi 4| < \delta$ then $-1\le \sin x \le 1$ and $-6 \le \sin x - 5 \le -4$ and $|\sin x - 5| < 6 < 7$.

Thus we have proven that for every $\delta$ there exist an $\epsilon> 0$ so that whenever $|x -\frac \pi 4| < \delta\implies |\sin x - 5| < \epsilon$.

What does that prove? Does it prove $\lim_{x\to \frac \pi 4} \sin x= 5$?

No it does not. It doesn't prove a damned thing because finding an $\epsilon>0$ so that $|x-\frac \pi 4| < \delta \implies |\sin x - 5|$, although certainly is possible, doesn't prove anything because .... it can always happen. It isn't interesting!

You say it is big but it's not wrong. I say it is not wrong but it isn't useful! You can't do anything with it.


Proving that for every $\delta$ there exists an $\epsilon > 0$ so that whenever $|x-a|< \delta \implies |f(x) - L|< \epsilon$ does NOT mean $\lim_{x\to a}f(x) = L$.

Now let's consider then opposite

Does $\lim_{x\to a}f(x) = L$ mean for every $\delta$ there exists an $\epsilon > 0$ so that whenever $|x-a|< \delta \implies |f(x)-L| < \epsilon$?

The second example is that if $f(x) = \frac 1x$ then $\lim_{x\to 1} f(x) = 1$. (You said the function is undefined at $x =1$. But $f(1) = \frac 11 = 1$. That most certainly IS defined.)

But it is not true that for every $\delta$ there exists an $\epsilon > 0$ so that whenever $|x-a|< \delta \implies |f(x)-L| < \epsilon$

If we take $\delta = 1.1$ we can't find any $\epsilon$ so that if $|x - 1| < \delta = 1.1$ then $|f(x)-1| < \epsilon$. For any $\epsilon$ we might consider $x= \frac 1{\epsilon + 2}$ and $0 < \frac 1{\epsilon + 2}=x< 1$ so $|x-1| < \delta$. But we $|f(x) -1|=|\epsilon + 2-1|=|\epsilon + 1| > \epsilon$ even though $|x - 1| < \delta$.


Result 2

$\lim_{x\to a}f(x) = L$ does NOT mean for every $\delta$ there exists an $\epsilon > 0$ so that whenever $|x-a|< \delta \implies |f(x)-L| < \epsilon$?


So what good is that condition?

The condition does NOT show us that limits exist.

And limits existing does NOT imply the condition.

So what good is the condition? What can we do with it?

And the answer is .... nothing really. I mean, it may or may not be true in and of its own right but it's not likely to be significant. It just is not interesting.

  • $\begingroup$ Great post, but not sure if I follow everything! Let me first clarify some things in your post. Should line 5 be $|x-4| < \delta$, not $x$? Second, your Result 1 referred to my post. In my post, I meant, the function $f(x)=\frac{1}{x}$, choosing $x=1$, is not defined when $\delta \ge 1$. Namely, $x-\delta$ would be less than 0. I have now updated it. $\endgroup$
    – user13985
    Feb 21, 2020 at 2:49
  • $\begingroup$ Yes it should be $|x-4|< \delta$. But what to you mean $x=1$ is not defined when $\delta > 1$? $x=1$ is always defined. Um.... so what if $x-\delta < 0$? That happens every time $x < \delta$. Why would you think anything is wrong with that? $\endgroup$
    – fleablood
    Feb 21, 2020 at 3:48
  1. Since sine is continuous, for any $a$, we should have a single limit, $L$, for each $a$. As Ilmari Karonen points out, the reversed definition does not succeed with a single value of $L$, it succeeds with arbitrary choices of $L$. When $L$ is allowed to be arbitrary, then the output of sine is simultaneously close to infinitely many non-close real numbers. This is absurd -- the output of sine can't be close to two non-close real number at any (much less every) real input.
  2. You are ignoring that the source example set $a = 1$ and $L = 1$. $f$ is continuous on $\Bbb{R} \smallsetminus \{0\}$, so is certainly continuous at $x = a = 1$. The value of the function is $1^{-1} = 1 = L$. So whatever definition of continuous you use, it has to hold for $a = L = 1$ for this $f$. The reverse definition fails to detect continuity here, so is not a definition of continuity.
  3. $\sin(1/x)$ is discontinuous at $x = 0$. The reversed definition fails to detect this discontinuity, so is not a definition of continuity.
  • $\begingroup$ For point #3, I'm actually using $delta$-$\epsilon$ to prove limit at $x=0$, not continuity. So, is there anything wrong now? $\endgroup$
    – user13985
    Feb 20, 2020 at 22:36
  • $\begingroup$ @user13985 : $\delta$-$\varepsilon$ is a definition of continuity. We have external definitions of continuity and discontinuity (removable, jump, infinite, essential, et al.). If your reverse definition fails to agree, it is not a definition of continuity. $\endgroup$ Feb 20, 2020 at 22:40
  • $\begingroup$ All my questions are regarding limit proof. How would point #3 violate limit definition? $\endgroup$
    – user13985
    Feb 20, 2020 at 22:42
  • $\begingroup$ A continuous function agrees with its limits at any point of continuity. If we show your definition fails to capture this property, then your definition is broken. $\endgroup$ Feb 20, 2020 at 22:43
  • 1
    $\begingroup$ @user13985 : Take a function, continuous at $a$, through the point $(a,L)$. We know that function agrees with its limit at $a$. If your definition of "limit" fails to produce that agreement, you do not have a working definition of limit. And vice versa. $\endgroup$ Feb 21, 2020 at 4:14

I think you need to understand the motivation behind the concept of limit of a function.

The concept of limit allows us to study local behavior of a function. The term "local behavior" means behavior in a certain neighborhood of a point. I hope you are aware of the term neighborhood. If you are not familiar then a neighborhood of a point $c$ is any open interval $I$ containing $c$. Now here comes the dilemma that a neighborhood $I$ of $c$ necessarily contains points other than $c$ and thus $I$ also acts as a neighborhood of another point $d$ with $d\neq c$. Then how does studying the behavior of a function in some neighborhood $I$ of $c$ remains related (or shall we say local) to $c$?

Well, to answer that question convince yourself of the obvious fact that if $c\neq d$ then we can find a neighborhood $I$ of $c$ and a neighborhood of $J$ of $d$ such that they have no points in common ie $I\cap J=\emptyset $. Further if the difference $|c-d|$ is small then we need to deal with smaller neighborhoods $I$ and $J$ to ensure $I\cap J=\emptyset$. Thus if we truly want to study the behavior of a function local to a point $c$ (and not local to another nearby point $d $) then we have to deal with arbitrarily small neighborhoods of $c$.

There is another catch here. We specifically do not want to study the behaviour of $f$ at $c$ precisely because it's trivial (just evaluate $f(c) $ and you are done). That brings us to the concept of deleted neighborhood. If $I$ is a neighborhood of $c$ then set $I\setminus\{c\} $ is a deleted neighborhood of $c$.

Thus we have the following problem :

Let $f$ be a real valued function defined in a certain deleted neighborhood of $c$. How does $f$ behave (in terms of trend of its values) in arbitrarily small deleted neighborhoods of $c$?

Some notations were invented to specify the above problem concisely and then the problem can be stated as

How do the values $f(x) $ behave as $x\to c$?

To answer this question we need to specify precisely the kind of behavior we are interested in. More specifically we are interested in knowing whether the values of $f(x) $ lie near some specific number $L$ (they may equal $L$ also) when we start considering all the values of $x$ lying in arbitrarily small deleted neighborhoods of $c$. If this happens to be the case then we say that the limit of $f$ at $c$ is $L$ or symbolically $\lim\limits _{x\to c} f(x) =L$.

Now comes the problem of making this statement as precise as possible and yet being reasonably useful. One important aspect is that if the values of $f$ are near $L$ then we have to ensure that they are not near $M$ for any $M\neq L$. As before this forces us consider the disjoint neighborhoods of $L$ and $M$. And since the difference $|L-M|$ can be arbitrarily small we need to consider arbitrarily small neighborhoods of $L$.

Another aspect is that we need to consider all points of neighborhood of $c$. This is to disallow the following kind of behavior: for every deleted neighborhood $I$ of $c$ there are some points in $I$ where values of $f$ are near $L$ and further there are some other points in $I$ where values of $f$ are not near $L$.

Finally we need to respect the following principle of local behavior:

If $f, g$ are two real valued functions and there exists a deleted neighborhood $I$ of $c$ such that $f(x) =g(x) $ for all $x\in I$ then their local behavior at $c$ must be same.

Taking into consideration these aspects we reach the modern definition of limiting behavior :

Let $f$ be a real valued function defined in a certain deleted neighborhood of $c$. A number $L$ is said to be the limit of $f$ at $c$ if for every neighborhood $J$ of $L$ there is a corresponding deleted neighborhood of $I$ of $c$ such that $f(I) \subseteq J$.

The positive numbers $\epsilon, \delta$ are used to quantify the size of neighborhoods $J$ and $I$ respectively.

It should come as a bit of a surprise that to deal with behavior of $f$ in arbitrarily small neighborhoods of $c$ we have to define a concept which needs arbitrary neighborhoods of $L$ instead of such neighborhoods of $c$. However there is no apparent paradox here as once we determine the suitable deleted neighborhood $I$ of $c$ it includes all smaller neighborhoods. So the definition does consider the values of $x$ in arbitrarily small deleted neighborhoods of $c$.

If we try to proceed according to your suggestion in question then we can't have a precise definition of limiting behavior. More formally your approach does not give a clear yes/no answer to the question: do values of $f$ lie near $L$ when $x\to c$?

If you think carefully you will also notice that it violates the principle of local behavior. Consider two functions $f, g$ defined via $$f(x) =1/x,x\neq 0,f(0)=0$$ and $$g(x) =1/x,x\in[1/2,2],g(x)=1 \text{ otherwise} $$ Clearly they have same values in the neighbourhood $(1/2,2)$ of $I$ and hence their limiting behavior as $x\to 1$ should be same. According to your approach let's choose $L=1,\delta=2$ and then we have no value of $\epsilon$ which works for $f$ but there are values of $\epsilon$ which work for $g$.

To sum up, given a function $f$ and a point $c$ under consideration we fix a specific behavior by giving a proposed limit $L$ and a margin of error $\epsilon$ and then try to figure out whether the function really behaves in this specific manner in some neighborhood of $c$ or not. If a suitable $\delta$ exists then $f$ does have that specific behavior otherwise it does not.


If I understand you correctly, you want to say that $ \lim \limits _ { x \to a } f ( x ) = L $ if and only if, for each $ \delta > 0 $, for some $ \epsilon > 0 $, whenever $ \lvert x - a \rvert < \delta $, then $ \lvert f ( x ) - L \rvert < \epsilon $. Correct? (Here, I've swapped the roles of $ \delta $ and $ \epsilon $ from their usual roles in the first place where each appears, but kept them in their usual roles in the second place where each appears.)

Consider the following two examples, one of which meets your definition but is not the correct limit, and one of which fails your definition but is the correct limit:

  1. $ f ( x ) = \cases { 0 & for \( x \ne 0 \) \\ 1 & for \( x = 0 \) } $, $ a = 0 $, $ L = 1 $;
  2. $ f ( x ) = 1 / x $, $ a = 1 $, $ L = 1 $.

In (1), given any $ \delta > 0 $, pick $ \epsilon = 2 $; then no matter what, $ \lvert f ( x ) - L \rvert < \epsilon $. Yet this is not a limit. In (2), given $ \delta = 1 $, pick any $ \epsilon > 0 $; then if $ x = 1 / ( \epsilon + 1 ) $, we have $ \lvert x - a \rvert < \delta $, but $ \lvert L - f ( x ) \rvert \geq \epsilon $. Yet this is a limit.

More fundamentally, you've made this about large errors instead of small ones. If you want to prove something continuous (by your proposed definition), just pick a big enough $ \epsilon $ to cover all nearby values of the function; that's what I did in (1). But you can be defeated if $ \delta $ is big enough to reach arbitrarily large values of the function; that's what I did in (2).

  • $\begingroup$ Yes, you caught my point. But, the standard definition just says "give me an epsilon, I give you back a delta". 1) It didn't indicate how large the epsilon is either. 2) I can make $\epsilon$ depend on $\delta$. $\delta(\epsilon)$ or, I can instead solve for $\epsilon^{-1})\delta)$. $\endgroup$
    – user13985
    Feb 20, 2020 at 23:26
  • $\begingroup$ 1) Neither definition says whether $\delta$ or $\epsilon$ is large or small, but your definition still suggests that they should be large, while the standard one suggests that they should be small. This is because a small $\epsilon$ makes it harder to prove that a limit exists, but a small $\delta$ makes it easier; while in your definition, a large $\delta$ makes it harder to prove that a limit exists, while a large $\epsilon$ makes it easier. ($\forall$ is supposed to make something harder to prove, while $\exists$ makes it easier; that's why I'm looking for harder/easier when I do.) $\endgroup$ Feb 20, 2020 at 23:36
  • $\begingroup$ $\epsilon$ may be as small as you like. There is not limit to how small it can be and by making $\delta$ depending on $\epsilon$ that assures there will never be a limit. If we solve $\epsilon$ in terms of $\delta$ then we will have no way of knowing if there is or is not a limit on it. And if there is a limit, then...It's useless... And before you ask, going the other way doesnt put a limit on the $\delta$. As the delta is preposition in: IF $|x-a|< \delta THEN |f(x)-L|<\epsilon$ and not the *CONCLUSION a limit is impossible. We just take delta smaller than we solved for. $\endgroup$
    – fleablood
    Feb 20, 2020 at 23:37
  • $\begingroup$ Sure, my definition would suggest it's large. My intuition has always been that, you have $x$ first, then map it to $y$, through $f$. Change in $x$ controls the change in $y$. If there's not $x$ defined, then, there would never be a $y$, let alone an error $\epsilon$. $\endgroup$
    – user13985
    Feb 20, 2020 at 23:39
  • $\begingroup$ Maybe this will be better: in the regular/standard definition, $|f(x) - L| < \epsilon$ whenever $|x-c|<\delta$. Within the $\delta$ interval, it doesn't say that epsilon will get smaller with delta together. In fact, within the interval, delta could be keep on decreasing, but maybe epsilon will decrease, then increase, then decrease again. $\endgroup$
    – user13985
    Feb 21, 2020 at 0:00

We say given $\epsilon>0$ first because there is a meaning it has, and a certain role it plays, that is different from that of $\delta.$ So let us look at what these symbols mean in this context.

The idea is to say that no matter how close we are required to get to a certain number by using the values of a function alone, we can always achieve this goal.

This is exactly what we want to state more precisely. The required degree of closeness is what we designate by a positive number $\epsilon.$ This is exactly why this quantity is always mentioned first -- because it is a goal, it is a requirement; it has to be given. The idea is that no matter how small this given positive quantity is, we can always get even closer than this to the limit via values of the function. We express this by saying that there is a point $x_0$ at least arbitrarily close to the domain of the function so that whenever we are sufficiently close (by $\delta$) to $x_0$ in the domain, then the values of the function are as close as was desired (by $\epsilon,$ that is) to $L.$

When you see exactly what this means and the roles that the symbols play, you see that they cannot be interchanged. One of the quantities is a specification we want to meet, hence it must be given, and we have no choice about it, namely $\epsilon>0.$ The other is one that we find in order to meet the given target, namely how close we need to be to $x_0$ in the domain in order to be $\epsilon$-close to $L$ in the codomain. This other quantity is what is called usually $\delta.$

  • 3
    $\begingroup$ If they have a justifiable reason, let the downvoter please explain. $\endgroup$
    – Allawonder
    Feb 21, 2020 at 18:23

I'm OP. The answer is that we want the tightest $\epsilon$ bound. Any $\epsilon$ that's bigger than the tightest bound may work, but that's not good enough. I call choosing $\epsilon$ first: "reverse engineering".

Let's give an example: $$f(x)=2x $$

I call x the input, $\delta$ input error. Call y the output, $\epsilon$ output error.

$$f(\delta)=2 \delta =\epsilon$$

If in the the case of derivative, $$f(x+\delta)=2(x+\delta)= 2x+2\delta=f(x)+2\delta$$

Graphically, $f(x+\delta)=f(x)+2\delta$ means that one unit $\delta$ increase in x direction results in $2\delta$ increase in $y$. So, relationship of the x-y errors is $2\delta=\epsilon$.

Q: Why given $\delta>0$ first, then choose $\epsilon$ doesn't work?

A: Because $\epsilon$ can be unbounded. Let's say we choose $\delta=3$., then is $\epsilon = 6$. But, does it have to be 6? No! It can be 7, 8, 9, 1000. In fact, any number bigger than 6 will do. So, choosing $\delta$ first leaves $\epsilon$ unbounded.

Choosing $\epsilon$ first is like reverse engineering. To me, changing $y$ first, and watch what happens to $x$ is weird. Because I'm used to x as the independent variable, and y as the dependent variable.

Q: Why does "reverse engineering" work? What's the intuition behind it?

A: The $\epsilon$-$\delta$ proof has two parts. First, "Given $\epsilon>0$, there exists a $\delta$." I call it the "proof conditions". Second, $|x-c|<\delta \implies |f(x)-L|<\epsilon$. I call it the "actual proof", this is forward engineering. The intuition is that, to make sure forward engineering works, we must make sure $\epsilon$ is not just any upper bound 10, 100, 1000. It's the smallest possible, $\epsilon=6$. So, we do reverse engineering by picking $\epsilon$ first.

Acknowledgement: I thank Arturo Magidin who discussed with me back and forth over the days. Great mentor. Thank you!

  • $\begingroup$ The example at the start seems to indicate that there should be relation between $\epsilon, \delta $. The kind of relation you are trying to establish with your example is handled by concept of derivative which deals with the problem of how fast/slow $f(x) $ changes as $x$ changes. The relation between $\epsilon, \delta $ is just "for every $\epsilon>0$ there is a $\delta>0$ such that.." $\endgroup$
    – Paramanand Singh
    Feb 26, 2020 at 8:28
  • $\begingroup$ Also you state that there are two parts in a limit proof. But these parts are highly interconnected. One does not find a $\delta$ and then proves that the implication is true. But rather one analyzes the desired implication and if necessary simplify it till the point it becomes obvious what value of $\delta$ will make it true. Also no unique value of $\delta$ is desired so we have great flexibility in analyzing and simplifying the implication. $\endgroup$
    – Paramanand Singh
    Feb 26, 2020 at 8:34
  • $\begingroup$ @ParamanandSingh You are right. I guess I was thinking about derivative, which isn't the same as a limit. I guess I'm getting limit and derivative mixed up. Taking off the green check until then. $\endgroup$
    – user13985
    Feb 27, 2020 at 0:13
  • $\begingroup$ @ParamanandSingh I agree with your second comment. I broke down the whole proof into two parts. I don't mean to say they are separated. By "One does not find a $\delta$ and then proves", do you mean that we have to keep on making $\delta$ smaller to get to the limit? If so, I agree. $\delta$ is only an interval which can be controlled to be however large or small. $\endgroup$
    – user13985
    Feb 27, 2020 at 2:19
  • 1
    $\begingroup$ The definition demands the existence of $\delta>0$ given an $\epsilon>0$. This does not mean that there is necessarily a functional or solvable relation between them. And there is no need to actually evaluate $\delta$. You should understand that existence is different from an explicit evaluation. Thus for example there exists a unique real number $x$ such that $x^5=x+16$ but we can't really express this $x$ as a finite expression consisting of rational numbers combined by arithmetic operations and radicals. $\endgroup$
    – Paramanand Singh
    Feb 28, 2020 at 0:49

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