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$$$\forall \epsilon>0:\exists \delta>0:\forall x\in\mathbb{R}:(0<\lvert x-x_0\rvert\leq\delta\implies \lvert f(x)-L\rvert\leq\epsilon)$$

$\rightarrow$ 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.


1 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.

  • $\begingroup$ 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 $$0<\lvert x-x_0\rvert<\delta\leq\Delta(\frac{\epsilon}{2})\implies\lvert f(x)-L\rvert<\frac{\epsilon}{2}\leq\epsilon$$ $\endgroup$ May 29, 2018 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, 2018 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.