TITLE: changing the order of the logical symbols $(\forall \epsilon) (\exists\delta)$ by $(\exists \delta)(\forall\epsilon) $ in limit definition
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.