Can anyone please give an example of why the following definition of $\displaystyle{\lim_{x \to a} f(x) =L}$ is NOT correct?:

$\forall$ $\delta >0$ $\exists$ $\epsilon>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$

I've been trying to solve this for a while, and I think it would give me a greater understanding of why the limit definition is what it is, because this alternative definition seems quite logical and similar to the real one, yet it supposedly shouldn't work.

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    $\begingroup$ Every function defined in a neighborhood of $a$ satisfies that property. Can you see why? $\endgroup$
    – anon
    Sep 8, 2012 at 20:11
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    $\begingroup$ Yes, you are correct. If f(x) is defined near a, then you can choose an epsilon as big as needed so that abs(f(x)-L)<e will be true for every L. Thanks a lot. $\endgroup$
    – Sebastian
    Sep 8, 2012 at 20:17
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    $\begingroup$ You've got $\varepsilon$ and $\delta$ reversed! Other than that, your definition is right. $\endgroup$ Sep 8, 2012 at 20:18
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    $\begingroup$ Michael Hardy: thanks, but I think anon got it right. There's more that having epsilon and delta reversed here. The quantifiers are reversed too, and I think that is a game changer. $\endgroup$
    – Sebastian
    Sep 8, 2012 at 20:21
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    $\begingroup$ @anon: even if $f$ is unbounded? To me, this statement basically says that $f$ is locally bounded (regardless of what $L$ might exactly be, so long as it's finite). $\endgroup$
    – tomasz
    Sep 9, 2012 at 0:01

5 Answers 5


I'm a little confused when you ask wether it is correct or not. Do you ask as opposed to the well known definition? That is, as opposed to

We say that $\lim\limits_{x\to a}f(x)=\mathscr L$ if for every $\epsilon >0$ there exists a $\delta >0$ such that, for all $x$, if $0<|x-a|<\delta$ then $|f(x)-\mathscr L|<\epsilon$.

If so, you can just see $\epsilon$ and $\delta$ are reversed.

However, suppose this definition was not given, and we want to define what we mean by limit. First, we clearly want to understand that we're concerned about what happens near $a$, but not at $a$. The idea of a "limit" near $a$ is then that of a value $\mathscr L$ a function $f(x)$ approaches when $x$ approaches $a$, So the idea we want to capture is that a function $f(x)$ has a number $\mathscr L$ as a limit when $x$ approaches $a$ if we can make $\mathscr L$ and $f(x)$ as near as we wish, by taking $x$ sufficiently near $a$. So when do we say that two number are near? We need to formalize our idea of proximity of numbers.

$\bf A$. So, we can associate to each $x,y\in \bf R$ the real number $d(x,y)=|x-y|$ and call it the distance from $x$ to $y$. Note that this distance has some properties we really want any notion of distance to have:

$(1)$ The distance is symmetric. $$d(x,y)=d(y,x)$$

$(2)$ It is always positive, unless $x=y$. That is, the distance of two numbers is zero if and only if they are the same number. $$d(x,y)>0 \text{ and } d(x,y)=0\iff x=y$$

$(3)$ The distance from a point $x$ to a point $y$ will always be less than or equal to the distance from $x$ to another $z$ plus the distance from that $z$ to $y$. We're just saying the shortest distance from $x$ to $y$ is precisely the straight line that joints them (this generalizes to higher dimensions). $$d(x,y)\leq d(x,z)+d(z,y)$$

$\bf B$. Now, consider this silly theorem:

Suppose $x,y\in \Bbb R$. Then, for every $\epsilon>0$, $d(x,y)<\epsilon$ if and only if $x=y$.

One direction is easy: if $x=y$, then clearly $d(x,y)=0<\epsilon$ for positive $\epsilon$. Now suppose that for any $\epsilon>0$, we have that $d(x,y)=0<\epsilon$. We aim to prove $x=y$. We will argue by contradiction. Since the distance is symmetric, we might assume $x<y$. Then $d(x,y)=|x-y|>0$. So $|x-y|$ is an $\epsilon>0$, which would mean that $|x-y|<|x-y|$, which is absurd. Thus, it must be that $x=y$.

$\bf C$. Now, we have a formal definition of the notion of near, and we can say that two numbers are near if $d(x,y)$ is small. In particular we just showed that if $x=y$, $d(x,y)$ is a smaller than any positive quantity given, since it is zero. It is but only logical to say that making $\mathscr L$ as near as we wish to $f(x)$ is making $|f(x)-\mathscr L|<\epsilon$ , no matter how small $\epsilon$ is, with $\epsilon<0$.

$\bf D$. We can now try and think about a degree of closeness. Given a number $\delta >0$, we can standarize and say that $x$ and $a$ are sufficiently near if $d(x,a)<\delta$. Think of it as the ruler we use to tell wether you can go in a rollercoaster ride or not. If $d(x,a)\geq \delta$, then $x$ is "bad" and we discard it.

$\bf E$. It is important to note this: we are concerned about making $f(x)$ and $\mathscr L$ close, and we want to succeed in doing so by making $x$ close to $a$. It is not our objective to make $x$ and $a$ close, but our means. So our definition must capture this: given a desired proximity of $f$ and $\mathscr L$, there must be a moment in which any $x$ close enough to $a$ will make $f(x)$ be close to $\mathscr L$. We don't forget that $x\neq a$, which only means $0<|x-a|$. So, let's try and write something down, considering what we discussed in $\bf B,C,D$.

We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any prescribed degree of closeness, making $x$ sufficiently close to $a$, but not equal to $a$, will imply that for all those $x$, $f(x)$ and $\mathscr L$ will be within that prescribed degree of closeness.

Applying $\bf B,C$, we can write

We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any $\epsilon >0$, making $x$ sufficiently close to $a$, but not equal to $a$, will imply that for all those $x$, $|f(x)-\mathscr L|<\epsilon$.

Using $\bf D$, we can write

We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any $\epsilon >0$, there is a $\delta$ such that for all $x\neq a$, $|x-a|<\delta$ implies $|f(x)-\mathscr L|<\epsilon$.

Finally, since $x\neq a$ is the same as $0<|x-a|$, we can be more succint and write.

We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any $\epsilon >0$, there is a $\delta$ such that for all $x$, $0<|x-a|<\delta$ implies $|f(x)-\mathscr L|<\epsilon$.

You can prove, or find proofs, that if the limit of $f$ exists for some $a$, then it is unique. And this is important. As others have pointed out, this definition, giving freedom on $d(f(x),\mathscr L)$,for example, makes limits lose their uniqueness.

Now, your definition clearly doesn't capture our original idea: It is somehow saying $f$ has $\mathscr L$ as a limit if for any prescribed degree of closeness $\delta$, there will exist some positive number $\epsilon$ such that making $d(x,a)<\delta$ will imply $d(f(x),\mathscr L)<\epsilon$. But this is giving us a lot of freedom on $d(f(x),\mathscr L)$, which we are most worried about. Of course we can make $x$ and $a$ as near as we wish, but the question is, can we make $f$ and $\mathscr L$ as close as we wish?

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    $\begingroup$ I think your last point is very important but perhaps not emphasized clearly enough. Let me see if I got it right: the definition that the OP suggested basically doesn't restrict $f(x)$ to get close to $L$ which is the main problem. For example we can be as close as we want to $a$ (for any $\delta$) but we could in fact keep choosing $\epsilon$s that get larger and larger...but what we want is $f(x)$ to get closer to $L$. Which the OPs definition doesn't capture. $\endgroup$ Jul 7, 2018 at 23:51
  • $\begingroup$ One of the best explanations I have read. One question. Point C. "It is but only logical to say that making ℒ as near as we wish to 𝑓(𝑥) is making |𝑓(𝑥)−ℒ|<𝜖 , no matter how small 𝜖 is, with 𝜖<0" I want to understand why $ \epsilon $ is shown as $<0$, and not $>0$. $\endgroup$ Mar 26 at 16:25

There are two problems with your "backwards" definition, which I'll illustrate with examples:

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

  2. Let $f(x) = 1/x$ (for $x \ne 0$, and let $f(0) = 0$, just to make $f(x)$ defined everywhere), and let $a = 1$. Then, for any $L$ (including $L = 1$), your definition fails for all $\delta \ge 1$, since for any $\epsilon$ we can choose $x=1/(L+\epsilon)$ if $L+\epsilon > 1$ and $x=1$ otherwise, so that in either case $f(x)-L \ge \epsilon$.

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    $\begingroup$ Thanks! I wasn't aware of the second problem with the backwards definition but it does makes sense. $\endgroup$
    – Sebastian
    Sep 8, 2012 at 21:09
  • $\begingroup$ I don't get it. What's wrong with $\epsilon = |L| + 2$? You found an $\epsilon$! They never said it has to how big or how small. $\endgroup$
    – user13985
    Feb 20, 2020 at 20:01
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    $\begingroup$ @user13985 "You found an ϵ!" But we shouldn't have! $\lim_{x\to a} \sin x \ne L$ so if if finding an $\epsilon$ had ANYTHING to do with taking limits we shouldn't have been able to find an $\epsilon$. But we did! So the reverse definition can't have anything to do with limits. ...All birds lay eggs. Is the reverse true. Lizard lays eggs. It failed. "but" you say "What's wrong with lizard eggs? You found eggs!". Yes, but finding eggs wasn't the point! Neither is finding the $\epsilon$. $\endgroup$
    – fleablood
    Feb 21, 2020 at 1:24
  • $\begingroup$ I don't understand the second problem... in order to refute the OP alternative definition shouldn't $\epsilon$ be chosen for each $\delta$? Why is $x$ being chosen instead? $\endgroup$
    – Javi
    Aug 18, 2020 at 4:35
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    $\begingroup$ @Javi: OP's definition is missing a quantifier for $x$, but in the usual definition of continuity the implication is intended to hold for all $x$ (with $\epsilon$ and $\delta$ independent of $x$; if they could vary depending on $x$, OP's definition would be even more broken). Thus, finding a single $x$ for which the implication does not hold is sufficient for a counterexample. $\endgroup$ Aug 18, 2020 at 8:31

Let's consider a counter example. Let's use your definition to prove that the limit of $x$, as $x$ tends towards 1, is 2. For all $\delta > 0$, we claim that there exists $\varepsilon_{\delta} > 0$ such that:

If $|x - 1| < \delta$ then $|x-2| <\varepsilon_{\delta}$.

We could define $\varepsilon_{\delta} := \delta + 2$. That seems to satisfy your definition.

  • $\begingroup$ Thanks! This is an effective and simple example. $\endgroup$
    – Sebastian
    Sep 8, 2012 at 21:16

Your condition can be translated into words as:

$f$ is bounded in every open ball around $a$

This is quite different from being continuous!

PS Not every function satisfies this condition, but the ones that don't are rather fierce. (In fact, in a strict sense, almost no function satisfies the condition, but the functions arising in analysis normally do.)

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    $\begingroup$ Thats a really interesting point. Do you mind expanding on why you think this is a very different notion to continuous? I'm having issues understanding continuous and right now from the definition I'd say that the best understanding I have of it is that for every $\epsilon>0$ there is some $\delta$ where we do not have any sudden jump that is larger than $\epsilon$. This seems to say that $f$ is $\epsilon$-bounded for every neighborhood of the limit point $a$. This seems identical to your description in words. Thus, do you mind clarifying the difference? That would be really valuable! $\endgroup$ Jul 7, 2018 at 21:41

The main issue with the definition that you suggested is that its trivially satisfied as other people pointed out. If you want to get $\delta$ close (really close) to $a$ then your definition says you need to find any $\epsilon$ (thats the problem! its too flexible). Trivially set $\epsilon = \infty$ (i.e. it should hold by the archimedian property i.e. just find a large enough number until $|f(x) - L| < \epsilon$). Then the the consequent of your implication will always hold as long as $f(x)$ is defined in the neighbourhood $\delta$. As Pedro pointed out brilliantly, the issue is that your definition provides to much freedom to choose $\epsilon$. In fact we intuitively want that as $\epsilon$ gets smaller and smaller that $x$ gets closer and closer to $a$ (or at least the largest delta gets smaller and smaller). This does hold true for the original definition of limits (though its not explicit in the definition) as shown in my answer here. Therefore to my understanding, your questions shows why people came up with the definition of limits the way they did, because:

  1. it's not clear how to define it cleanly starting first with $\delta$'s and have the property that we get closer and closer to the value limit value $L$ as we get closer to the limit point $a$.
  2. as your question shows, if we start with $\delta$ first, then $\epsilon$ can't be just specified with a single quantifier. Choosing existence is too flexible and "every" also is too restrictive. $\forall \delta $, $\forall \epsilon$, is obviously too intense. So every function value in the neighbourhood around $a$ must be arbitrarily close to the limit point? That just can't be true for most sensible notions of distance unless everything maps to one point.
  3. therefore, instead we proceed to specify a restriction on $\epsilon$ and say we want to be able to be arbitrarily close to the limit $L$ (since we want to approach the limit L, thats nearly the whole point!). Now we just need to find a set of points around $a$ that approach it, and if its possible to have a definition that as we want to be closer to the limit $L$ that the points get closer to $a$, the better! The original definition does satisfy that and since it uses friendly quantifiers it's more compact (and elegant I guess) and its easier to sue and it has the properties we want. Perhaps the second point that I really find important is not explicit but would you prefer to have a really long complicated definition just to make that explicit? I the answer must be no for mathematicians and poof thats why the definition is the way it is!

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