Switching "For all $\epsilon$, there exists $\delta$, ..." to "For all $\delta$, there exists $\epsilon$, ..." in the definition of function's limit Formal accepted definition: For all $\epsilon >0$, there exist $\delta >0$ such that if $0<\vert x-a\vert<\delta$, then $\vert f(x)-L\vert<\epsilon$.
The other statement: For all $\delta >0$, there exist $\epsilon >0$ such that if $0<\vert x-a\vert<\delta$, then $\vert f(x)-L\vert<\epsilon$. 
I know that in quantifier logic, switching quantifiers might change the statement itself, but both of the above statements seem non-contradicting to me so must it be the same?
 A: Consider the function $f : \mathbb{R} \to \mathbb{R}$ given by
$$
f(x) = \begin{cases} 0, & x \leq 0; \\1, & x > 0. \end{cases}
$$
The limit of $f(x)$ as $x$ tends to $0$ is not defined, or in other words, it is not true that for every $\epsilon > 0$ there exists $\delta > 0$ such that $|f(x) - L| < \epsilon$ for all $x$ such that $0 < |x| < \delta$. In particular, take $\epsilon = 1/2$. No matter what $L$ you choose, you cannot find such a $\delta$ (show this!).
On the other hand, look at what happens when you switch the epsilons and deltas. It is certainly true that for every $\delta > 0$ you can find an $\epsilon > 0$ such that $|f(x) - L| < \epsilon$ for all $0 < |x| < \delta$, and it doesn't even matter what $L$ you pick! Why? Let's say $L = 0$. Observe that taking $\epsilon = 2$ works (any $\epsilon > 1$ works, in fact). If you pick some other $L$, say $L = 3.5$, then choose $\epsilon = 4$. See how taking larger and larger values of $\epsilon$ allows me to make this statement true for any value of $L$ and any value of $\delta$, for this function $f$. Do you see why this fails to tell me anything about the existence of a limit? (Hint: in the second definition we are not talking about the behaviour of $f$ close to $L$. Also see @A.Γ.'s comments above.)
A: An example: let $a \in \mathbb R, \delta >0, I:=(a- \delta,a+ \delta)$ and let $f:I \to \mathbb R$ be a function. Then we have:
$f$ satisfies the second definition $ \iff f$ is bounded.
