In the formal statement of the epsilon-delta definition of a limit, what is the scope of the existential quantifier?
I.e. in this statement where $x$ is a real number:
$$ \forall \epsilon > 0, \exists \delta > 0, (0 < \mid x - c \mid < \delta) \implies (\mid f(x)-L \mid < \epsilon) $$
Does the '$\exists \delta$$'$ quantifier apply to the open sentence
$$ (0 < \mid x - c \mid < \delta) $$
Or does it apply to the whole conditional
$$ (0 < \mid x - c \mid < \delta) \implies (\mid f(x)-L \mid < \epsilon) $$
If it is the second case (in that it quantifies the entire conditional), then isn't the statment trivially satisfied by choosing any $\delta > \mid x - c \mid$? Because then the antecedent of the conditional is always false and therefore the conditional is always true, implying the existence of the limit $L$?
This makes me think the existence quantifier only applies to the antecedent, but I'm slightly confused because I assume the universal quantifier applies to the entire sentence (and neither of the scope's of the quantifiers appear to be explicitly shown with parenthesis in the definitions I have seen. So I don't know what the logic is to determine whether it applies to the entire sentence, or just the immediate sentence after.)
You are giving the definition of $\lim_{x \to c} f(x)=L$, I suppose, so $c$ and $L$ are pre-given values. $x$ however is also a variable and needs a quantifier: for each precision $\varepsilon$ we need a precision $\delta$ such that whenever it is the case that $x$ and $c$ are not equal but $\delta$-close we always have that $f(x)$ is $\varepsilon$-close to $L$, or formally:
$$\forall \varepsilon >0: \left(\exists \delta >0: \left(\forall x \in \Bbb R: ((0<|x-c|< \delta) \implies (|f(x) - L| < \varepsilon)\right)\right)$$
so we cannot take $\delta$ depending on $x$ but the implication statement should hold for all $x$.