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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.

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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.

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  • $\begingroup$ Would this work: $\rightarrow$ we'll choose $\frac{\epsilon}{2}$ and $\delta=\frac{\Delta(\frac{\epsilon}{2})}{2}$ such that $$0<\lvert x-x_0\rvert\leq\delta<\Delta(\frac{\epsilon}{2})\implies \lvert f(x)-L\rvert\leq\frac{\epsilon}{2}<\epsilon$$ $\leftarrow$ we'll choose the same and $$0<\lvert x-x_0\rvert<\delta\leq\Delta(\frac{\epsilon}{2})\implies\lvert f(x)-L\rvert<\frac{\epsilon}{2}\leq\epsilon$$ $\endgroup$ – Slavik Egorov May 29 '18 at 16:27
  • $\begingroup$ @SlavikEgorov That works. But I'll caution to be a little more careful with your notation: $\Delta$ means something different in the two directions (when you're doing the forward direction it means the $\delta$ which satisfies the limit definition, the backwards direction the $\delta$ which satisfies the $\leq$ condition), so I'd use two different names. $\endgroup$ – BallBoy May 29 '18 at 16:31

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