The informal intuition for the limit of a function is this:

What is the value of the function $f$ as $x$ gets infinitely close to $c$?

How on earth does this monster

$$ \lim_{x \to c} f(x) = L \iff (\forall \varepsilon > 0)(\exists \ \delta > 0) (\forall x \in D)(0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon)$$ capture our intuition?

I want the answer to explain and motivate every bit of this formal statement. Remember: I am able to parse the statement, I just don't see how it captures our informal intuition.

  • $\begingroup$ This is a common problem when students first see $\epsilon-\delta$ definitions. They are a lot more subtle than they look. $\endgroup$ Oct 20, 2015 at 21:22
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    $\begingroup$ @uniquesolution The title may look similar, but the question is a different one: it's about the dependency of $\delta$ on $\epsilon$. The (only) answer to that question does not answer the present one. $\endgroup$
    – user147263
    Oct 20, 2015 at 23:07
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    $\begingroup$ @uniquesolution I don't think the question is a duplicate. This question asks about the intuition behind the definition, while the linked question asks about how $\delta$ depends on $\epsilon$. In my opinion, this question is more fundamental (in the sense that someone first encountering limits will more likely ask this one, not the linked one). $\endgroup$
    – 5xum
    Oct 20, 2015 at 23:45
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    $\begingroup$ The informal definition of limit is not what you mention, bur rather somewhat more contrived. By $f(x) \to L$ as $x \to a$ we informally mean that "values of $f(x)$ can be made arbitrarily close to $L$ for all values of $x$ sufficiently close to (but not equal to) $a$". The phrases "arbitrarily close" and "sufficiently close" are informal and they are formalized by using $\epsilon - \delta$. An "informal definition" does not mean "non-rigorous definition", rather it means "avoiding the use of too many symbols". $\endgroup$
    – Paramanand Singh
    Oct 21, 2015 at 4:21
  • $\begingroup$ math.stackexchange.com/questions/1156419/… $\endgroup$
    – user117644
    Nov 22, 2015 at 11:26

5 Answers 5


The definition can be read, in human words:

For every positive $\epsilon$, there exists such a positive $\delta$ that if $|x-c|$ is smaller than $\delta$, but larger than $0$, then $|f(x) - L|$ is smaller than $\epsilon$.

First of all, lets's get the double inequality out of the way. Basically, $0<|x-c|$ is just saying that $|x-c|$ is not equal to $0$ (since it can't be negative), and saying that is just saying that $x-c$ cannot be $0$, or in other words, that $x$ is not allowed to equal $c$.

Now to a non-mathematician, that still makes very little sense, but take into account that $|a-b|$ is really the distance between numbers $a$ and $b$.

So, we can translate the definition into

For every positive $\epsilon$, there exists such a positive $\delta$ that if $x$ and $c$ are two distinct numbers and the distance between them is smaller than $\delta$, then the distance between $f(x)$ and $L$ is smaller than $\epsilon$.

But that still does not ring quite "natural" But what does the "for all $\epsilon$, there exists a $\delta$" in the beginning mean anyway? Well it means that whatever $\epsilon$ you give me, I can find a $\delta$ such that the condition will be true. So:

No matter what $\epsilon$ you choose, I can find such a positive $\delta$ that whenever you take any $x$ near (but not equal to) $c$ that is less than $\delta$ away from $c$ which are closer together than $\delta$, $f(x)$ will be closer than $\epsilon$ away from $L$.

Getting warmer to something readable? Well, let's get rid of the variables even further:

No matter how close you want $f(x)$ to be to $L$, I can tell you how close to $c$ you need to pick your $x$, and if you pick your $x$ that closely, then I can guarantee that $f(x)$ will be as close to $L$ as you originally wanted it to be.

This is very similar to what MPW wrote in comments:

$f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $c$

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    $\begingroup$ I think you've not only shown that the intuition is preserved, but that in fact the two statements are homotopic :) Very nice. $\endgroup$
    – pjs36
    Oct 20, 2015 at 21:16
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    $\begingroup$ Nice, +1. One further step is usually something along the lines of "$f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $c$" $\endgroup$
    – MPW
    Oct 20, 2015 at 21:18
  • $\begingroup$ @MPW I added that into the answer. $\endgroup$
    – 5xum
    Oct 20, 2015 at 21:20
  • $\begingroup$ $f$ need not be defined at the point $c$, which is why the inequality posted by the OP is $0<|x-c|<\delta$, and not just $|x-c|<\delta$. Your first three yellow boxes actually require $f$ to be defined at $c$, which is just a tiny loss of generality. $\endgroup$ Oct 20, 2015 at 22:20
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    $\begingroup$ I'd really like to hear the explanation from the downvoters... what about the answer is bad enough to warrant a downvote? $\endgroup$
    – 5xum
    Oct 23, 2015 at 15:06

The definition capture not an open question as

''What is the value of the function $f$ as $x$ gets infinitely close to $c$''

but the exact statement:

$f$ can be as close as we want to $L$ if $x$ is sufficiently close to $c$.

If you think to this you can see that the ''monster'' works well.

  • $\begingroup$ This actually did answer my question. Thank you! $\endgroup$
    – user132181
    Oct 23, 2015 at 11:20

It means that we can make $f(x)$ take on a value as close as we like to $L$ (closer than any positive $\epsilon$ that we choose) by bringing $x$ near enough to $c$ (nearer than the positive $\delta$ which is asserted to exist in the definition).


Imagine that you want to be $\epsilon$ close to to limit $L$, than the definition just says that you can pick arbitrary any $x$ from some $\delta$ neighborhood of $c$ and with $f(x)$ you are $\epsilon$ close.


Intuitively it makes sense if you read the right-hand side as saying "it is possible to get $f(x)$ as close to $L$ as you like by demanding only that $x$ get very near $c$". If the right-hand expression is always true, then the limit on the left will be true. This does not imply that $f(c)=L$, so $f$ is not necessarily continuous [that would be something like $\lim_{x->c}f(x)=f(c)$]. Finally, to gain an intuitive feeling for your monster take a function for which the left-hand limit fails [eg., $f(x)=0$ if $x<0$, $f(x)=1$ otherwise], then [with $c=0, L=1$] try to get the right-hand side to work [ie., given $\epsilon=0.1$ try to find a $\delta$ which works]. Do this and your monster will make sense.


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