The intuitive definition for $\lim\limits_{x \to a} f(x) = L$ is the value of $f ( x )$ can be made arbitrarily close to $L$ by making $x$ sufficiently close, but not equal to, $a$ . I can easily understand this ,but for the (ε, δ)-definition of limit:For every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ implies | f(x) − L | < ε. Oh, god, I cannot understand it completely.

As it is the formal definition of limit, I think it should be precise but somewhat should include the mean of the above intuitive definition, so as for "$f ( x )$ can be made arbitrarily close to $L$" in the intuitive definition correspond to “For every real ε > 0,| f(x) − L | < ε” in the formal definition, it's fine! but does “making $x$ sufficiently close, but not equal to, $a$ .” correspond to “there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ” ? This is the point I cannot understand, because I am not sure if “there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ” shows "x close enough, but not equal, to $a$".

The other question is: does the biggest δ also get smaller as ε is getting smaller? Why? (Exclude the case when f(x) is a constant function.)

Why do we need the formal definition of a limit? Does the intuitive definition have some flaw?

P.S. Thank you everyone, but I must declare I only have some basic knowledge of limit, I only started to learn calculus a few days ago.

  • $\begingroup$ Non-standard analysis provides a framework that may be closer to your intuition. $\endgroup$
    – polmath
    Sep 25, 2014 at 15:00
  • $\begingroup$ $\delta$ does not have to get smaller as $\epsilon$ gets smaller. You just have to know that no matter how small $\epsilon$ is, you can find your $\delta$. In some cases, $\delta$ might get bigger as $\epsilon$ shrinks. $\endgroup$
    – graydad
    Sep 25, 2014 at 15:03
  • 3
    $\begingroup$ We need the formal definition of a limit to make precise what we mean by the phrases used in the intuitive definition. For example "making $x$ sufficiently close, but not equal to $a$". If you use the definition to prove that a limit is $L$, you need the precision of the mathematical version. $\endgroup$ Sep 25, 2014 at 15:08
  • 3
    $\begingroup$ Have you looked at some of the posts listed in the "Related" column? For example, Interpretation of $\epsilon$-$\delta$ limit definition, I Need Help Understanding the Formal Definition of A Limit, and Need help in understanding definition of limit of a function. $\endgroup$ Sep 25, 2014 at 15:27
  • 1
    $\begingroup$ You might consider modeling the definition as a game. There is a challenger who gives you an $\epsilon$, saying in effect, "Can you guarantee that $f$ stays this close to $L$?" You answer by giving the challenger a $\delta$ and saying "If you stay within this distance of $a,$ the challenge is met." The limit exists when you are always able to satisfy the challenger. $\endgroup$ Sep 25, 2014 at 15:44

4 Answers 4


The rationale behind the concept of limit is exception handling: We have a function $f:\>\Omega\to {\mathbb R}$ defined on some set $\Omega$, and we are given a "place" $a$ which belongs to $\Omega$, or at least is "adherent" to $\Omega$. Therefore the function $f$ may or may not be defined at $a$. But we observe (e.g., by letting Mathematica draw the graph of $f$) that "when $x$ is near $a$ then $f(x)$ is near a particular value $\eta$". If that is the case we'd like to tell this to other people by writing $\lim_{x\to a} f(x)=\eta$.

Now we need a formal definition for such a fact. Under what circumstances would a value $\eta$ qualify as limit of $f(x)$ when $x\to a$? The answer is simple: If defining $$f(a):=\eta\tag{1}$$ (resp. overriding the given definition of $f(a)$ by $(1)$) would make $f$ continuous at $a$.

Now appeal to the definition of continuity: A function is continuous at $a$, if, given any tolerance $\epsilon>0$ we can guarantee $|f(x)-f(a)|<\epsilon$ by choosing $|x-a|$ "sufficiently small", i.e., smaller than a certain allowance $\delta>0$, which will depend on the given tolerance $\epsilon$.

  • $\begingroup$ "place" and "adherent" keywords turn on me an intuitive idea of what the concept of limit means. Maybe there are various ideas about the concept of limit. $\endgroup$
    – baudo2048
    Aug 23, 2017 at 8:22
  • $\begingroup$ "place" and "adherent" keywords turn on me an intuitive idea of what the (formal) concept of limit means. But, maybe, there are various ideas that model/represent the concept of limit (limit of what?). So a problem come out, how to formalize those ideas? Set theory should be the foundation for the formal definition. The good old set nevertheless is overloaded with such "spatial" ("metric"?) concepts! So my question is: Can formal description of limit concept be given out of a metric space (and, in general, out of a topological space)? Does exist concept of limit without concept of continuity? $\endgroup$
    – baudo2048
    Aug 23, 2017 at 8:45
  • 1
    $\begingroup$ @baudo2048: The question asked for an intuitive description of the limit concept. The exact definition is in all textbooks. I'd say that continuity is the simpler concept, and should be introduced first. But this is a matter of taste. $\endgroup$ Aug 23, 2017 at 9:10

First, notice that

$f(x)$ can be made arbitrarily close to $L$ by choosing $x$ close enough to $a$

is equivalent to

No matter how close you want $f(x)$ and $L$ to be, you can limit the distance between $x$ and $a$ such that $f(x)$ and $L$ will be as close as you wanted.

Let's give names ($\varepsilon$ and $\delta$) to the two "close" distances:

For any maximum distance $\varepsilon > 0$ between $f(x)$ and $L$, there is a maximum distance $\delta > 0$ between $x$ and $a$, such that whenever $x$ and $a$ are closer than $\delta$, $f(x)$ and $L$ are closer than $\varepsilon$.

Removing redundant words, we get:

For any $\varepsilon > 0$, there is a $\delta > 0$, such that whenever $x$ and $a$ are closer than $\delta$, $f(x)$ and $L$ are closer than $\varepsilon$.

And using $|a - b|$ instead of "distance between $a$ and $b$", we get:

For any $\varepsilon > 0$, there is a $\delta > 0$, such that whenever $|x - a| < \delta$, $|f(x) - L| < \varepsilon$.


The $\varepsilon$-$\delta$-definition of the limit can be read like this:
for any neighbourhood of $L$, no matter how small (points with $|y-L|<\varepsilon$), there is a neighbourhood of $a$ (the points with $0<|x-a|<\delta$), such that $f$ maps the entire neighbourhood into the neighbourhood of $L$ (that is $|f(x)-L|<\varepsilon$).
This captures the idea of the intuitive definition. The intuitive way of seeing this makes $\delta$ smaller as $\varepsilon$ gets smaller, since being closer to $L$ is a stronger condition and is satisfied by fewer points around $a$. This is just not required by the definition.
If there is a $\delta>0$ that satisfies $|f(x)-L|<\varepsilon$ for all $0<|x-a|<\delta$ then this also true for all $0<\delta'<\delta$, you can always choose a different $\delta$ that is smaller than all others you have seen before, and imagine $\delta$ being monotonically decreasing. The important point is that $\delta$ is non-zero, that there always is a neighbourhood of $a$ such that all points of it get mapped arbitrarily close to $L$ by $f$.

  • $\begingroup$ There are many functions $f:\mathbb{R}\to\mathbb{R}$ where $f(x)$ is defined via a limit for every point $x$, derivatives are the most common example I think. If you then take a limit of $f$ you have a limit of limits, it is cases like that most need a formal definition because intuition no longer helps. $\endgroup$
    – planckh
    Sep 30, 2014 at 20:03

As you get closer and closer to the x you are approaching, does the function get closer and closer to the limit? Notice there's two things getting smaller here.

In other words, x is arbitrarily close to a, is f(x) arbitrarily close to L?

The point is we have to choose the $\delta$, such that f(x) is as close to L as we need.

Here's an example,

$$\lim_{x \to 2} x^2 $$ and $$ |f(x) - 4| = |x^2 - 4| = |x-2||x+2|. $$

The first thing you should notice is we have a $|x + 2|$ term that us unwanted, so we need to bound it under $\epsilon$. Because we have $0 < |x - 2| < \delta$, this term will take itself out if $\delta < 1$.

Choosing $\delta < 1$, we have $ |x+2| < \delta+4 < \delta + 4\delta = 5\delta $ because $ |x - 2| < \delta $, so we can choose $\delta < min(1,\frac{\epsilon}{5})$ then,

$$ |f(x) - 4| = |x-2||x+2| < |x + 2| < 5\delta \leq 5\frac{\epsilon}{5} = \epsilon.$$

See how I compare $\delta$ to $\epsilon$? I already had $\delta$ in terms of $\epsilon$, so it was trivial to ensure the limit was smaller than $\epsilon$.

  • $\begingroup$ @user1485853 I included an example proof using epsilon/delta. Let me know if anything confuses you. $\endgroup$
    – user109879
    Sep 25, 2014 at 15:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .