Prove/disprove: $\lim_{x\to x_0}f(x)=L\iff \leq \delta$ and $\leq \epsilon$

Question

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.

Answer 1

Your attempt for the $\implies$ direction falls short when you try to work out the details. You'd have to show $|x-x_0|\leq \delta \implies |x-x_0|< \delta \implies|f(x)-L|< \epsilon \implies|f(x)zL|\leq \epsilon$. You have the second implication by definition and the third implication by the argument you presented, but not the first.

Instead, you want to think about $\delta$ as a function of $\epsilon$ and change the parameters as follows. Let $\Delta(\epsilon)$ give the $\delta$ which satisfies the limit definition for $\epsilon$. To satisfy the given condition, choose a smaller $\delta$, e.g., $\frac{\Delta(\epsilon)}2$.

I'll leave you to work out the details, and the other direction is similar. Let me know if you need more help.