Let $f:\mathbb{R}\to\mathbb{R},\quad x_0\in \mathbb{R}$ be given.
Prove/disprove: $\lim_{x\to x_0}f(x)=L\iff$$$\forall \epsilon>0:\exists \delta>0:\forall x\in\mathbb{R}:(0<\lvert x-x_0\rvert\leq\delta\implies \lvert f(x)-L\rvert\leq\epsilon)$$
attempt
$\rightarrow$ in the limit definition the $<\delta$ and $<\epsilon$ imply $\leq \delta$ and $\leq \epsilon$ since $<$ are in particular $\leq$.
$\leftarrow$ this is where my intuition is telling me that since we look at $\{f(x)\lvert -\epsilon+L\leq f(x)\leq \epsilon+L\}$ we can find a smaller error $\epsilon'>0$ with a smaller $\delta(\epsilon')>0$ so that this direction is proved. However, if that is the case, I'm not sure how to put in out mathematically.