# Assumptions when proving limit using epsilon delta definition

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.

• 'we could not guarantee that $|x^2+x-20| < 10 * 1$, if we take $\delta$ to be 1 ', Why? Commented Aug 14, 2023 at 9:04
• @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? Commented Aug 14, 2023 at 9:34
• Does this answer your question? Does there exist an epsilon-delta proof of the limit for all polynomials?
– user1211588
Commented Oct 8, 2023 at 21:36

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.
• 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. Commented Aug 15, 2023 at 5:27