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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?

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  • $\begingroup$ 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 $\endgroup$ – B. Chinaski May 20 '16 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. $\endgroup$ – MathematicsStudent1122 May 20 '16 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 May 20 '16 at 8:37
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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.

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  • $\begingroup$ 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. $\endgroup$ – Lost definition May 23 '16 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. $\endgroup$ – Zelos Malum May 23 '16 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 Oct 13 '16 at 2:39
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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:

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

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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 $P(\epsilon)$.

While

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.

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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

$$|f(x_1)-f(x_2)|=|f(x_1)-L+L-f(x_2)|\le|f(x_1)-L|+|f(x_2)-L|$$

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,

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

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.

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

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

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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