Is this an accurate and correct way to represent the epsilon-delta definition in full?
$$\lim_{x \rightarrow a} f(x) = L \Leftrightarrow \forall \varepsilon > 0,\ \exists \delta > 0,\ \forall x,\ (0< |x-a| < \delta) \rightarrow (|f(x) - L| < \varepsilon)$$
(even if it's just a little bit off or inaccurate I'd like to know what the correct version is).
I was also unsure what it looks like negated because of the if-and-only-if piece. So far I have it to
$$\lim_{x \rightarrow a} f(x) \neq L \Leftrightarrow \exists \varepsilon > 0,\ \forall \delta > 0,\ \exists x,\ (0< |x-a| < \delta) \land (|f(x) - L| \geq \varepsilon)$$
What's the correct way to state the definition as well as its negation? How do you negate the if-and-only-if piece here? Because $a \Leftrightarrow b = (a \rightarrow b) \land (b \rightarrow a)$ so negating it would mean $(a \not\rightarrow b) \lor (b \not\rightarrow a)$
The definition, valid for $a,L \in \mathbb{R}$, seems to be ok.
And also the negation should be ok, note indeed that the if and only if part is out of the statement since it is a part of the definition, that is
$$\lim_{x \rightarrow a} f(x) = L \Leftrightarrow P$$
$$\lim_{x \rightarrow a} f(x) \neq L \Leftrightarrow \lnot P$$
maybe at the end of the negation (and then also for the definition), following the suggestion by Fimpellizieri, we could simply take
$$\lim_{x \rightarrow a} f(x) \neq L \Leftrightarrow \exists \varepsilon > 0,\ \forall \delta > 0,\ \exists x,\ (0< |x-a| < \delta), (|f(x) - L| \ge \varepsilon)$$
but it seems to be not essential and equivalent.