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I have a question. If I want to prove some limits, I use the $\delta-\epsilon$ definition. And I have to choose my $\delta := \min\{\delta_1,\delta_2,...,\delta_k\}$ to be sufficient. Why does it work? I have found one usefull answer, but I don't understand a very important place. So if we have the precise definition of limit.

$\lvert x-a \rvert<\delta$ then $\lvert f(x)-L \rvert< \epsilon$

which we might denote $P(\delta)$, regarding $f,a,L$ and $\epsilon$ as given/known.

If the condition $P(\delta)$ is true for some $\delta>0$ and if $0<\delta'<\delta$, then $P(\delta')$ is also true, $\textbf{because its hypothesis is logically more strict}$. I don't understand, why does it work? For example if $\lvert x-a \rvert<\delta$ and $0<\delta<\delta'$ its clearly that it will work for $\delta'$ because its greater, but why will it work for $\delta'$ if $0<\delta'<\delta$?

Thank you for help!

Gary
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mathguruu
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  • For example, if we assume $|x - a| < 0.01 = \delta'$, then we are assured $|x - a| < \delta = 0.1$. (Think carefully about what your intuition tells you, which is not correct: $|x - a| < \delta$ is a hypothesis here, not a conclusion. (A link to the other question might help....) – Andrew D. Hwang Jun 15 '22 at 10:31
  • Guessing the referenced question was this? – Andrew D. Hwang Jun 15 '22 at 10:37
  • @AndrewD.Hwang thank you for your answer! Aaah, I understood it. So if $|x-a|<0.01=\delta'$ it's clearly that $|x-a|<\delta$, BUT it does not guarantee, that for this $\delta$ we have $|f(x)-L|<\epsilon$. Did I understood it right? Link for this question - https://math.stackexchange.com/questions/966789/why-do-we-need-min-to-choose-delta – mathguruu Jun 15 '22 at 10:38
  • @AndrewD.Hwang yes:) – mathguruu Jun 15 '22 at 10:38
  • @mathguruu: It seems you've got it now. – Vasili Jun 15 '22 at 10:52
  • That's essentially right, though to be completely correct we need to be explicit about the relationship between $\epsilon$ and $\delta'$. As in the linked answer, I'd prefer to think of it as "making $\delta$ smaller makes the hypothesis stronger". – Andrew D. Hwang Jun 16 '22 at 10:20

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If we have $\delta:=min\{\delta_1,...,\delta_k\}$, then by definition, $\delta \leq \delta_i$ for each $i$. Thus if $|x-a| \leq \delta$, we have $|x-a| \leq \delta_i$ for each $i$. So this definition allows us to take the most stringent of conditions.

Peng Hao
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