Let’s say I have a limit for example $\lim_{x\to 4}x^2 + x -11= 9$ and I want to prove this limit using the epsilon-delta definition. So then $|x-4| < \delta$ and $|x^2+x-20| < \epsilon$

Now $|x+5||x-4| < \epsilon$

Assuming $\delta \le 1$

$ -1 < x-4 < 1$

$8< x+5 < 10$

$|x+5| < 10$

$|x-4||x+5| < |x-4|*10 < \epsilon$

$|x-4| < \epsilon/10$

$\delta \le \epsilon/10$

and also

$\delta \le1$

thus $\delta = \min({1, \epsilon/10})$

Using our choices of delta

$|x-4| < \delta \le \epsilon/10$ and also $|x-4|< \delta \le 1$ So now to prove it

$|x^2+x-20| = |x^2+x-20|$

$|x^2+x-20| = |x+5||x-4|$

$|x^2+x-20| < 10 * \epsilon/10$

and finally

$|x^2+x-20| < \epsilon$

$|x^2 + x -11 - 9| < \epsilon$

I get the process and why we do this, it's to show that we can find a value $\delta$ such that whenever x is in the range of $a-\delta$ and $a+\delta$ $\implies$ f(x) will be in the range of $L-\epsilon$ to $L+\epsilon$ but what I'm unsure of are the assumptions

We assumed that $\delta < 1$ and proved that if that is the case, then $|x^2+x-20| < \epsilon$ but my question is, what if it was the case that $\epsilon/10 > 1$, wouldn't taking $\delta = 1$ break the definition of our limit, as we could not guarantee that $|x^2+x-20| < 10 * 1$, if we take $\delta$ to be 1

Would we be unable to prove the limit via the definition?

and if possible could someone provide a case in which we wouldn't be able to prove a limit using the definition (due to the fact it doesn't exit) because I was not able to find one online.

  • 2
    $\begingroup$ 'we could not guarantee that $|x^2+x-20| < 10 * 1$, if we take $\delta$ to be 1 ', Why? $\endgroup$ Commented Aug 14, 2023 at 9:04
  • $\begingroup$ @geetha290krm I just graphed it, you are right, sorry. But in that case does that mean that we only know that our limit exists if our value that we pick for $\epsilon$ is one such that $\delta < 1$ Otherwise the only thing we have proven is that the function is smaller than 10 provided that $\delta$ = 1, so the only way we can truly say the limit exists is if we find a value $\epsilon$ such that $\delta$ is smaller than 1 or? $\endgroup$
    – Dan Lupu
    Commented Aug 14, 2023 at 9:34
  • $\begingroup$ Does this answer your question? Does there exist an epsilon-delta proof of the limit for all polynomials? $\endgroup$
    – user1211588
    Commented Oct 8, 2023 at 21:36

1 Answer 1


The $\varepsilon-\delta$ limit definition requires that for any $\varepsilon$ that we pick, we are able to find a suitable $\delta$. It doesn't work in reverse - we don't need to be able to find $\varepsilon$s that work for specific $\delta$s, and it doesn't require that the $\delta$s are unique to each $\varepsilon$.

One way to think about it is like a two-player game. Player 1 picks a value of $\varepsilon$, and Player 2 has to bound $f(x)$ within $\pm \varepsilon$ of the limit by naming a value of $\delta$. If Player 2 has a strategy to always be able to meet Player 1's challenge, then the limit is correct. If there's a value of $\varepsilon$ that Player 1 can name that prevents Player 2 from naming a suitable value of $\delta$, then the limit is wrong. So in this example, if Player 1 names a ridiculously large value of $\varepsilon$, Player 2 can safely set $\delta = 1$ and not even think about it. It's only if Player 1 names a small value for $\varepsilon$ that Player 2 needs to pick a value of $\delta$ that directly relates to it.

  • $\begingroup$ So, if I understand correctly, the statement we have proven is if epsilon is a value greater than 10 or equal, f(x) will never be above epsilon + a or a-epsilon as long as |x-a| < 1, but if epsilon is smaller than 10, then we can always find a value delta such that f(x) is bounded between epsilon + a and a -epsilon by the formula, delta = epsilon/10 $\endgroup$
    – Dan Lupu
    Commented Aug 15, 2023 at 5:06
  • 1
    $\begingroup$ That's pretty much it. And from that, you know that you can always choose a suitable $\delta$, which means you've proven the limit. $\endgroup$
    – ConMan
    Commented Aug 15, 2023 at 5:27
  • $\begingroup$ Thank you so much, your analogy is very clear! $\endgroup$
    – Dan Lupu
    Commented Aug 15, 2023 at 5:32

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