changing the order of the logical symbols $(\forall \epsilon) (\exists\delta)$ by $(\exists \delta)(\forall\epsilon) $ in limit definition 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 $\delta&gt;0$ for all $\epsilon&gt;0$ such that, for all x, whenever $ 0&lt;|x-a|&lt;\delta$ then $|f(x) - L|&lt;\epsilon$


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?
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.
As the $\epsilon$ bound is independent of the $\delta$, for any $\epsilon&gt;0$, 

$$|f(x)-L|&lt;\epsilon$$

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