Some time ago a professor told the class, which I was in, to analyze why this definition of limit is not good (or if it is a good definition to argument why):

There exists a $\delta>0$ for all $\epsilon>0$ such that, for all x, whenever $ 0<|x-a|<\delta$ then $|f(x) - L|<\epsilon$

Or, in other words, changing the order of the logical symbols $(\forall \epsilon) (\exists\delta)$ by $(\exists \delta)(\forall\epsilon) $ in the epsilon-delta limit definition implies a bad functioning definition of limit?

Recently this question has returned from the universe of non answered question and it has brought another doubt. Is there some counterexample that can bring a contradiction to this bad definition?

When I try to respond this question I can't find a counterexample since everything I think is linked with the original limit definition. I am still trying to find sufficient small $\delta$ for a arbitrary small $\epsilon$. But when I enunciate the above definition, intuitively something sounds different.

Can anyone help?

  • $\begingroup$ A function satisfies your requirement if and only if $f$ is bounded in any interval of center $a$. In this case, $L$ can be any real number $\endgroup$ Commented May 20, 2016 at 4:16
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    $\begingroup$ The problem is, there isn't a $\delta$ which works for all $\epsilon$, unless your function is locally constant. $\endgroup$ Commented May 20, 2016 at 4:42
  • $\begingroup$ When you write "$\forall \epsilon > 0, \exists \delta > 0$ you mean that for each $\epsilon$ there corresponds a $\delta$ and for different values of $\epsilon$ the value of $\delta$ may be different. If we reverse the order i.e. $\exists \delta > 0, \forall \epsilon > 0$ it means that we have existence of one single $\delta$ for all the $\epsilon$'s. This is not what we wish to say when defining limit. $\endgroup$
    – Paramanand Singh
    Commented May 20, 2016 at 8:37

5 Answers 5


Let's analyse this logically, because what else do we do in mathematics?

There exists $\delta$ for all $\epsilon$

This entails that for some $\delta$ we have that it is independed of our $\epsilon$. That means if $\epsilon=1$ or $\epsilon=10^8$ we have the same $\delta$, which is bad because it can be suddenly at any difference. For example

$$f(x)=\sin x$$ we can have $\epsilon=2$ and clearly there exists a $\delta$ satesfying the criteria but equally clearly is it that it does encompasses all all points, so if we let $x\to 0$ we have it converges to all values in $[-1,1]$.

The key here is that $(\exists\delta)(\forall\epsilon)$ implies that if $\delta$ is thought of as a function of $\epsilon$ then $\delta(\epsilon)=C$, that is with respect to $\epsilon$ it's constant for that given value.

  • $\begingroup$ Thanks Zelos Malum. This example was very clear. But this reopen another question in my mind. When we have to prove that a limit does not exist, we use the universal and existential quantifiers in the switched order. So, can I affirm that in order to prove that a limit does not exist I have to prove that $\epsilon$ and $\delta$ are independent? And to prove that a limit exist, do I have to prove that they are dependent? Do you agree with this? I extend this question to all people that kindly answer my question. $\endgroup$ Commented May 23, 2016 at 1:38
  • $\begingroup$ When you prove that a limit exist then you prove that $\delta(\epsilon)$ is indeed a function such that the criteria is satesfied, when you prove a limit does NOT exist then you show that such a function cannot exist by demonstrating no matter what function you have, you can always show it does not satesfy the criterias given. $\endgroup$ Commented May 23, 2016 at 3:19
  • $\begingroup$ The negation of $(\forall\epsilon)(\exists\delta)P$ is $(\exists\epsilon)(\forall\delta)\neg P$. That's why the quantifiers look switched when you disprove the existence of a limit. $\endgroup$
    – arkeet
    Commented Oct 13, 2016 at 2:39

Consider the difference between $~\forall \epsilon~\exists \delta~( \epsilon < \delta)~$ and $~\exists \delta~\forall \epsilon~(\epsilon <\delta)~$.   Are these equivalent statements?

The first statement says: "For every number there is a number that it is less than."   This is true; because there is no maximum real number.

The second statement says: "there is some number such that every number is less than it."   This is false; because there is no maximum real number.

So we see that changing the order of universal and existential quantifiers may have significant impact on the meaning of a statement.

It is much the same with:

$$~\forall \epsilon~\exists \delta~\forall x~\Big(\big( g(x)<\delta\big) \;\to\; \big(h(x)<\epsilon\big)\Big)\quad\require{cancel}\cancel\iff \quad \exists \delta~\forall \epsilon ~\forall x\Big(\big( g(x)<\delta\big) \;\to\; \big(h(x)<\epsilon\big)\Big)$$


I don't completely agree with the other answers, I'll explain why.

Let's have a closer look at:

There exists a $\delta>0$ for all $\epsilon>0$ such that, for all x, whenever $ 0<|x-a|<\delta$ then $|f(x) - L|<\epsilon$

Of course, there is the immediate answer, that permuting quantifiers changes the meaning, as has already been explained by several answers. However, while I know it's bad, I have seen teachers do that in casual language, and maybe it's a cultural bias because english is not ma native language, but to me,

There exists a $\delta>0$ for all $\epsilon>0$

does not mean the same as

There exists a $\delta>0$ such that for all $\epsilon>0$

But it may mean

For all $\epsilon>0$ there exists a $\delta>0$

That is, I could accept this part of the sentence, though it's very error-prone.

I would prefer symbolic quantifiers $\forall$ and $\exists$, for which the order is more formally defined. Then, $(\forall\epsilon)(\exists\delta)$ and $(\exists\delta)(\forall\epsilon)$ have clearly a different meaning.

I have another concern, the lack of comma between for all $\epsilon>0$ and such that. Usually, in mathematics,

for all $\epsilon>0$ such that $P(\epsilon)$

Means for all $\epsilon>0$ that also satisfies $P(\epsilon)$.


for all $\epsilon>0$, $P(\epsilon)$

Means that for all $\epsilon>0$, $P(\epsilon)$ is true.

All in all, I would say the sentence is not correctly formed, because I would tend to interpret it as

( For all $\epsilon>0$ such that, for all x, whenever $ 0<|x-a|<\delta$ then $|f(x) - L|<\epsilon$ ), ( there exists a $\delta>0$ )

That is, it would make $\delta$ appear in the quantifier before its definition.

Notice that adding parens would likely clarify the intent, though it would probably look ugly:

( There exists a $\delta>0$ ) ( for all $\epsilon>0$ ) such that, for all x, whenever $ 0<|x-a|<\delta$ then $|f(x) - L|<\epsilon$

Maybe it's a dumb interpretation, and I would appreciate the comments of english speaking people on this. If it's that dumb, I will likely remove the answer altogether.


To continue off of what Zelos has said, the problem is that your choice of $\delta$ is independent of $\epsilon$. The problem is not so much that this $\delta$ works for $\epsilon=10^{1000}$ and $\epsilon=10^{-1000}$, but that it works for all $\epsilon$ smaller than even this.

Let's break it down: this $\delta$ means we're working in the interval $(x-\delta,x+\delta)$. But in this interval, for every $\epsilon$, no matter how small, we have $|f(x)-L|<\epsilon$. In this sense, this means that in the interval $(x-\delta,x+\delta)$, $f(x)$ is very close to the constant $L$, in fact the only function that satisfies this condition is a (locally) constant function!

To see this, let's assume otherwise, say that $x_1,x_2$ are two distinct points in $(x-\delta,x+\delta)$ and suppose that $f(x_1)\ne f(x_2)$. Let $d=|f(x_1)-f(x_2)|>0$, since we assume that $f(x_1)\ne f(x_2)$. However, note that


Since $x_1,x_2\in(x-\delta,x+\delta)$, this means by your definition that

$$|f(x_1)-L|<\frac{d}{2}$$ $$|f(x_2)-L|<\frac{d}{2}$$

so in totality,


Hence we get $|f(x_1)-f(x_2)|<|f(x_1)-f(x_2)|$, a clear contradiction.

So what does this tell us? Swapping $(\forall\epsilon)$ and $(\exists\delta)$ means that the limit of a function, in your $\epsilon-\delta$ definition, doesn't actually exist if the function isn't locally constant in some interval. Since non constant functions have proven to be occasionally useful, we traditionally use the quantifiers the other way around.


As the $\epsilon$ bound is independent of the $\delta$, for any $\epsilon>0$,


therefore $f(x)=L$ for all $0<|x-a|<\delta$. That is, $f(x)$ is a constant function on $0<|x-a|<\delta$.


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