How to rephrase these delta epsilon inequalities? When we say that $|f(x) - L| &lt; \epsilon$ can this be restated as $L - \epsilon &lt; f(x) &lt; L+\epsilon$? (removing absolute value to make this easier to understand)

If so I don't know how to restate $0 &lt; |x - c| &lt; \delta$ without the absolute value. because there's a $0 &lt;$ 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 &lt; x &lt; c+\delta $
No. It is equivalent to$$-\varepsilon+L&lt;f(x)&lt;\varepsilon+L.$$
By the definition of a limit, you know that for every $\epsilon &gt; 0$ there exists a $\delta &gt; 0$ such that for every $x \in D$ ($x \neq c$) exists $0 &lt; | x - c| &lt; \delta$, then $|f(x) - L| &lt; \epsilon$.

Because you know that $\delta &gt; 0$, then you could rewrite $$0 &lt; | x - c| &lt; \delta$$ as $$-\delta &lt; 0 &lt; | x - c| &lt; \delta$$

and from here to get:
$$-\delta &lt; x - c &lt; \delta$$