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The epsilon delta definition requires there to be a delta for all epsilon but at https://openstax.org/books/calculus-volume-1/pages/2-5-the-precise-definition-of-a-limit example 2.41 it says:

Prove $\lim\limits_{x \to 2} x^2 = 4$

Without loss of generality, assume $\epsilon \leq 4$ (since $\delta \leq 2 - \sqrt{4 - \epsilon}$), this is allowed because if we can find $\delta>0$ that “works” for $\epsilon \leq 4$, then it will “work” for any $\epsilon>4$ as well. Keep in mind that, although it is always okay to put an upper bound on $\epsilon$, it is never okay to put a lower bound (other than zero) on $\epsilon$.

I don't understand why a delta for a restricted range of epsilon implies there exist a delta for all epsilon as said above.

Paramanand Singh
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  • There is a counterpart. If one $\delta$ works for some $\epsilon $ then a smaller $\delta$ also works for the same $\epsilon $. These are self evident from the nature of inequalities given in definition of limit. – Paramanand Singh Jul 15 '20 at 02:31

2 Answers2

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Suppose that for each $\varepsilon\in(0,4)$ there is a $\delta>0$ such that$$|x-2|<\delta\implies|x^2-4|<\varepsilon.\tag1$$Now, take $\varepsilon>0$. You want to prove that there is a $\delta>0$ such that $(1)$ holds. If $\varepsilon<4$, you already know that such a $\delta$ exists. If $\varepsilon>4$, take $\delta>0$ such that $|x-2|<\delta\implies|x^2-4|<3$. Then, since $3<\varepsilon$, you know that $(1)$ holds for this $\delta$.

José Carlos Santos
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The definition of a limit says:

for every (meaning: for arbitrarily small) positive epsilon there exists delta (implied: small enough), such that...

So we are only interested in narrowing the epsilon range and seeing whether appropriate deltas exist for smaller and smaller epsilons.

If some delta 'works' for some specified epsilon, we are no longer interested in bigger values of epsilon – the same delta satisfies the condition for them.

Explicitly, if for some positive $\varepsilon_1$ and $\delta_1$ we have that $$x\in(p-\delta_1,p+\delta_1) \implies f(x)\in (q-\varepsilon_1, q+\varepsilon_1)$$ then for any $\varepsilon_2 > \varepsilon_1$ we also have $$x\in(p-\delta_1,p+\delta_1) \implies f(x)\in (q-\varepsilon_2, q+\varepsilon_2)$$ because $$(q-\varepsilon_1, q+\varepsilon_1) \subset (q-\varepsilon_2, q+\varepsilon_2).$$

CiaPan
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  • why does for all imply small. And what does small mean - it is subjective which contradicts the purpose of a precise definition of a limit –  Jul 19 '20 at 12:41
  • is it true that $f(x) \notin (q-\epsilon_2,q-\epsilon_1) \cup (q+\epsilon_1,q+\epsilon_2)$ if not why not? –  Jul 19 '20 at 12:50
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    @user716881 It implies 'small', because if the conjecture is satisfied for any specific value of epsilon, it automatically holds for all greater values – but not for smaller. Hence what we actually need to prove is that the conjecture holds for every smaller value than any chosen one. That is: for all 'arbitrarily small'. – CiaPan Jul 20 '20 at 12:02
  • Thanks, how obout my second comment –  Jul 22 '20 at 14:08
  • @user716881 Isn't this statement: $f(x)\in (q-\varepsilon_1, q+\varepsilon_1)$ enough to find out the answer? – CiaPan Jul 22 '20 at 14:15
  • Yes, I was just asking for my own knowledge –  Jul 22 '20 at 14:20
  • @user716881 The set you described is disjoint with the set I quoted. So if the value belongs to latter, it can't belong to the former. – CiaPan Jul 22 '20 at 14:29
  • $(q-\varepsilon_2,q-\varepsilon_1) \cup (q+\varepsilon_1,q+\varepsilon_2) \subset (q-\varepsilon_2, q+\varepsilon_2)$ so if $f(x) \in (q-\varepsilon_2, q+\varepsilon_2) \implies f(x) \in (q-\varepsilon_2,q-\varepsilon_1) \cup (q+\varepsilon_1,q+\varepsilon_2)$ however there is no $f(x)$ shown to be in $(q-\varepsilon_2,q-\varepsilon_1) \cup (q+\varepsilon_1,q+\varepsilon_2)$ –  Jul 22 '20 at 14:34