When we say that $|f(x) - L| < \epsilon$ can this be restated as $L - \epsilon < f(x) < L+\epsilon$? (removing absolute value to make this easier to understand)
If so I don't know how to restate $0 < |x - c| < \delta$ without the absolute value. because there's a $0 <$ on the end and I don't know how to incorporate it.
I don't think I can ignore the $0$ and just say $c-\delta < x < c+\delta $
On
By the definition of a limit, you know that for every $\epsilon > 0$ there exists a $\delta > 0$ such that for every $x \in D$ ($x \neq c$) exists $0 < | x - c| < \delta$, then $|f(x) - L| < \epsilon$.
Because you know that $\delta > 0$, then you could rewrite $$0 < | x - c| < \delta$$ as $$-\delta < 0 < | x - c| < \delta$$
and from here to get: $$-\delta < x - c < \delta$$
No. It is equivalent to$$-\varepsilon+L<f(x)<\varepsilon+L.$$