I'm a mathematics tutor trying to get a better understanding of the theory of epsilon-delta limit proofs.

I can prove linear and constant epsilon delta proofs easily, and I understand the proof form and definitions, but nonlinear ones stump me.

For example, take Problem 4., Page 3 here:

$$ \lim_{x \rightarrow 2} {x^2 + x - 2} = 4 $$

The paper works through it as follows: \begin{align} |f(x)-L| < \epsilon &\implies |(x^2+x-2) - 4| < \epsilon \\ &\implies |(x^2+x-6)| < \epsilon \\ &\implies|x+3||x-2| < \epsilon \\ &\implies |x-2| < \frac{\epsilon}{|x+3|} \end{align} Now, this is the point where I would add $$ \text{let} \ \delta = \frac{\epsilon}{|x+3|} $$ However, no source I've seen does that. This is where I get a little confused. The epsilon-delta definition means that the expression $$ \lim_{x\rightarrow c }f(x) = L $$ is equivalent to $$ \forall\epsilon >0 \ \exists\delta>0 \ s.t. 0 < |x-c| < \delta \implies |f(x) - L|<\epsilon. $$ Now, the definition has no qualifications for $x$. $x$ is just the input of the function. However, the paper goes on to say "In general delta must be in terms of epsilon only, without any extra variables." Why? The proof, when done "forward", goes through just fine for $\epsilon: \epsilon(\delta, x)$. And intuitively, this makes sense: for some parts of $f(x)$, the limits on epsilon will be different. However, every source I've seen finds limits on $|x+3|$ in some region and uses a constant in its place. Why must this be done? Why not leave $|x+3|$ in the denominator and be done with it?

  • $\begingroup$ Maybe better to write the definition as $$\forall\epsilon >0 \ \exists\delta>0\ \forall x [0 < |x-c| < \delta \implies |f(x) - L|<\epsilon].$$If you do not put the quanifier on $x$ in there, then in fact you cannot tell whether $x$ depends on $\epsilon$ or $\epsilon$ depends on $x$. $\endgroup$
    – GEdgar
    Commented Jan 27, 2020 at 18:28
  • $\begingroup$ @MarkViola So, are you saying that $\delta$ can depend on $x$? $\endgroup$
    – bjorn93
    Commented Jan 27, 2020 at 18:31
  • $\begingroup$ @bjorn93 I was claiming that $x$ does not depend on $\delta$, as you claimed it could. $\endgroup$
    – Mark Viola
    Commented Jan 27, 2020 at 18:38
  • 1
    $\begingroup$ The linked note by Lin has some merit in terms of explaining how one might find a suitable $\delta$, but it is most certainly not a model for how to set out a proof. The implications $|f(x)-L|<\epsilon\implies |x-x_0|<\epsilon$ are all the wrong way! So they are strictly for rough working, not for the final proof. $\endgroup$
    – almagest
    Commented Jan 27, 2020 at 18:42
  • $\begingroup$ @MarkViola What I meant is that $0<|x-a|<\delta$ makes $x$ dependent on $\delta$ by requiring it to be in a certain $\delta$-region around $a$. Perhaps it's not clear this is what I meant. $\endgroup$
    – bjorn93
    Commented Jan 27, 2020 at 18:43

5 Answers 5


If you can show that

$$ \forall\epsilon >0 \ \exists\delta>0 \ s.t. 0 < |x-c| < \delta \implies \delta = \frac{\epsilon}{|x+3|}, $$

then you might be able to use $\delta = \frac{\epsilon}{|x+3|}$ in your proof. Can you prove the implication?

The piece above is a broad hint about where you are going wrong. Here is a more explicit explanation:

You can think of the definition of a limit as a game you play against a perfectly clever opponent.

Your opponent moves first by selecting $\epsilon.$ They can choose any positive number whatsoever, tiny, huge, or in between.

Next, it's your turn; you get to select $\delta.$ You can choose any $\delta$ you like as long as it's positive.

Now it's your opponent's move again. They get to choose $x,$ but this time the choice is restricted to a punctured neighborhood of $c$ that you defined when you chose $\delta$; namely, they have to choose $x$ such that $0 < \lvert x - c \rvert < \delta.$

Now if $\lvert f(x) - L \rvert \geq \epsilon,$ you lose.

But to do an actual proof, you need to describe a perfect strategy so that your opponent can never defeat you. In effect, your formula for $\delta$ is a program for a limit-game-playing robot that plays your part of the game. Making your task even more difficult, your opponent will be able to know the program before the game starts. So if there is any flaw in the program, your opponent can exploit it and win. Flaws include values of $\delta$ that let the opponent set $x$ so that $\lvert f(x) - L \rvert \geq \epsilon,$ as well as cases where your rule doesn't produce a definite answer.

So let's say your "program" for the robot is $$ \delta = \frac{\epsilon}{|x+3|}.$$

Your opponent chooses $\epsilon = 10^{-100}.$ Now it's your move. What does your robot choose for $\delta$?

Your rule for $\delta$ requires the robot to know $x,$ which is a number your opponent has not selected yet. So your robot has no way to compute $\delta.$ Poor robot. You lose.

  • $\begingroup$ This response is both explicit and intuitive. Thanks so much! $\endgroup$ Commented Jan 28, 2020 at 14:52
  • $\begingroup$ So, $\delta$ must be the same for any $x$ and, therefore, $\delta$ itself can not depend on $x$. Is my understanding correct? $\endgroup$
    – eanmos
    Commented Oct 24, 2020 at 15:43
  • $\begingroup$ @eanmos Yes, I think that is a correct understanding. A typical delta-epsilon definition even has a part that is read, "for some $\delta,$ for all $x$," meaning one $\delta$ that is then the same one for all $x.$ $\endgroup$
    – David K
    Commented Oct 24, 2020 at 19:02
  • $\begingroup$ Appreciate you help, thank you. $\endgroup$
    – eanmos
    Commented Oct 25, 2020 at 4:10

The point is that you are not writting the complete definition of limit. It should be:

$$\forall \varepsilon>0\ \exists\delta>0; \forall x (\textrm{in the domain of}\ f); 0<|x-c|<\delta\Rightarrow |f(x)-L|<\varepsilon.$$

Notice there is a $\forall x$ such that $0<|x-c|<\delta$. This means you must find a delta in such a way that every $x$ which satisfies the condition $0<|x-c|<\delta$ will also satisfy $|f(x)-L|<\varepsilon$.

Geometrically, $0<|x-c|<\delta$ represents an interval centered in $c$ with radius $\delta$. The definition says that for $\varepsilon>0$ fixed there exists such a region which is mapped inside the interval centered in $L$ with radius $\varepsilon$.

  • $\begingroup$ To me, it seems that the definition you have provided proves continuity for all $x$ in the domain of $f(x)$. To prove a limit, I'm not particularly concerned about the entire domain, just a neighborhood around a point. I don't see how this extended definition changes my initial question. What am I misunderstanding here? Thanks! $\endgroup$ Commented Jan 27, 2020 at 18:40
  • $\begingroup$ Nope, the above definition only tells you that $\lim_{x\to c} f(x)=L$. The point $c$ is fixed, $x$ is varying inside the interval $(c-\delta, c+\delta)$. When I wrote $\forall x (\textrm{in the domain of}\ f); 0<|x-c|<\delta$ you must read: for all $x$ in the domain of $f$ which satisfies $0<|x-c|<\delta$, I don't mean "for all $x$ in the domain of $f$'' without further restriction. $\endgroup$
    – PtF
    Commented Jan 27, 2020 at 21:00

Many answers have emphasized what the definition of limit you've given means, i.e. that "$ \forall \epsilon \; \exists \delta$ ..." means that delta must be determinable from epsilon, but it seems none have answered the primary question of why this must be the definition to appropriately capture the notion of a limit. I'll try to address this. First, to established our intuition, the original definition ensures that no matter how small a region around $L$ we choose, we can find some neighborhood of $c$ on which $f(x)$ is always within that small region.

Let's consider what the definition according to your suggested interpretation might look like; that is, for a function $f:U \to \mathbb{R}$ ($U \subset \mathbb{R}$ being the domain of $f$) suppose we define $$\lim_{x \to c} f(x) = L \iff \left( \forall \epsilon>0 \; \exists\,\delta_\epsilon:U \backslash \{c\} \to \mathbb{R}_+ \text{ s.t. } |x-c| < \delta_\epsilon(x) \implies |f(x)-L| < \epsilon \right).$$ There's nothing self-contradictory about this proposed definition at the outset-- one can certainly write this down and choose to work with it. However, it utterly fails to capture what a limit should mean (or much of anything, for that matter), the primary reason being that $|x-c| < \delta_\epsilon(x)$ need not be satisfied by any (let alone all) $x$ sufficiently close to $c$ for arbitrary functions $\delta_\epsilon$. A limit should tell us what $f$ is doing everywhere around $c$, and the condition stipulated by this definition says nothing about those $x$ near $c$ that don't satisfy $|x-c| < \delta_\epsilon(x)$. What's more, this definition makes limits (very) nonunique.

To see these objections more clearly, consider the prototypical step function $$f(x) = \begin{cases} 1, & x \geq 0 \\ 0, & x < 0 \end{cases}$$ I expect we can agree that the limit $\lim_{x \to 0} f(x)$ should not exist, i.e. there is no one number that $f$ tends to "approach" as $x \to 0$. However, given $\epsilon > 0$, consider the function $$\delta_\epsilon(x) = \begin{cases} 2|x|, & x > 0 \\ |x|/2, & x < 0 \end{cases}$$ Then the statement that $|x-0| < \delta_\epsilon(x) \implies |f(x)-1| < \epsilon$ is true, since $|x-0| < \delta_\epsilon(x)$ is only satisfied for $x>0$, where $f$ is identically equal to $1$. That is, according to our proposed definition, $\lim_{x \to 0} f(x) = 1$! A very simple modification also demonstrates that $\lim_{x \to 0} f(x) = 0$. But the situation is even (much) worse than this! By choosing $\delta_\epsilon$ such that $|x-c| < \delta_\epsilon(x)$ is never satisfied (say $\delta_\epsilon(x) = |x-c|/2$), the definition is shown to be vacuously satisfied for any $L$ whatsoever!

We're therefore forced to the following conclusion: allowing $\delta$ to depend on $x$ as well simply breaks the definition's meaning entirely. Placing the necessary restrictions on the functions $\delta_\epsilon$ to make the above definition reasonable will ultimately give you the standard definition.

  • 1
    $\begingroup$ This is a great comment and clarifies a lot of remaining questions that I had. Thanks so much! $\endgroup$ Commented Jan 28, 2020 at 14:54

There are a few logical points to make here.

First, the very meaning of the statement

$\forall \epsilon > 0$ $\exists \delta > 0$ such that $P(\epsilon,\delta)$

is that once $\epsilon$ has been chosen, you can find one specific value of $\delta$, so that when you plug the chosen value of $\epsilon$ and the found value of $\delta$ into the statement $P(\epsilon,\delta)$, you make that statement true. If there are any further quantified variables inside the statement $P(\epsilon,\delta)$, then the $\delta$ is not allowed to depend on those variables.

Now let's look a little deeper. Your statement of the definition of limit has a small flaw: the variable $x$ has not been quantified. Let me add in the appropriate quantification for $x$.

$\forall \epsilon > 0$ $\exists \delta > 0$ such that $\underbrace{\forall x \in \mathbb R, \text{if $|x-a|<\delta$ then $|f(x)-f(a)| < \epsilon$}}_{P(\epsilon,\delta)}$

The fact that the quantified variable $x$ occurs inside the statement $P(\epsilon,\delta)$ makes it very clear: $\delta$ is not allowed to depend on $x$.


It is perhaps more informative to understand limits through the lense of continuity, instead of continuity through limits. In this post:

What is the intuition behind uniform continuity?

I discuss the idea of how that continuity can be best considered as the property which makes it possible to approximate the value of a function at a given input by suitably approximating the input - the property that makes it so the value that your calculator gives you when using functions can be relied on as having some semblance of meaning even if it is not, strictly speaking, the actual value of that function at the given point. That is, if we want to know the value of the function at $x_0$, and we have some tolerance $\epsilon$ we want to know it within (this is a $\pm$ tolerance, not a percentage or proportional tolerance), i.e. we demand

$$|f(x) - f(x_0)| < \epsilon$$

where $x$ is the point we're using to approximate the target $x_0$, then we want to find out how much we need to bring $x$ within of $x_0$ to achieve $\epsilon$ of accuracy, that is, find $\delta$ as a function of $\epsilon$, so that if we make

$$|x - x_0| < \delta(\epsilon)$$

then we are guaranteed at least $\epsilon$'s worth of tolerance in the function output. We are choosing $\epsilon$ first because our goal is to approximate the output of $f$, not to approximate $x_0$ with $x$. Approximating $x_0$ is the means to that end, so $\delta$ comes secondarily.

This idea carries over to limits. The limit

$$L = \lim_{x \rightarrow x_0} f(x)$$

is the value that $f$ "should have" at the point $x_0$ in order to be continuous there, i.e. it is the value $L$ which makes the function defined by

$$f^{*}_{x_0}(x) := \begin{cases} L,\ \mbox{if $x = x_0$}\\ f(x),\ \mbox{if $x \ne x_0$} \end{cases}$$

where $\mathrm{dom}(f^{*}_{x_0}) := \mathrm{dom}(f) \cup \{ x_0 \}$, continuous at $x = x_0$. And the same reasoning applies as before. Note that in this understanding of limits, their use to build the remainder of calculus (e.g. derivatives, integrals) is quite evident, since we are essentially using them to "fill in" the point where we'd like to evaluate at (e.g. a zero-size increment of the difference quotient) but can't, directly. This makes that "filling in" idea explicit right there in the definition of limit.


You must log in to answer this question.

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