I just need a quick clarification on this epsilon-delta epsilon proof, I'm not sure if the entire question is needed or not, for now i'll include the relevant part.

RTP: $\lim_{x\to5}(3x^2-1)=74$

Let $\epsilon_2$ = min{$\epsilon,72$}

And take $\delta = $min{ $5-\sqrt{25-{\epsilon_2/3}},-5+\sqrt{25+{\epsilon_2/3}}$ }

Then for every x, the expression



$-\delta\lt x-5 \lt \delta$

$-5+\sqrt{25-{\epsilon_2/3}}\lt x-5 \lt -5+\sqrt{25+{\epsilon_2/3}}$

My question is how are they able to take both values of $\delta$ in forming the last inequality, can you always do this? And is it okay to define a $\epsilon_2$ in order to avoid undefined $\delta$ as they have done.

Proof Cont.

$\sqrt{25-{\epsilon_2/3}}\lt x \lt \sqrt{25+{\epsilon_2/3}}$

${25-{\epsilon_2/3}}\lt x^2 \lt {25+{\epsilon_2/3}}$

$-{\epsilon_2}\lt 3x^2-75 \lt {\epsilon_2}$

$|3x^2-75| \lt {\epsilon_2} \le \epsilon$

As required.

  • $\begingroup$ Usually you find these from experience more than anything :) $\endgroup$ – Zelos Malum May 2 '17 at 13:21
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    $\begingroup$ Can you show more of the question and proof? These sorts of things are usually "reverse-engineered" depending on the needs of the proof. $\endgroup$ – MPW May 2 '17 at 13:24
  • $\begingroup$ Added the proof/question @MPW $\endgroup$ – Samson May 2 '17 at 13:35
  • $\begingroup$ Yes? @StackTD . $\endgroup$ – Samson May 2 '17 at 13:54
  • $\begingroup$ The proof above has serious shortcomings. If it from some textbook, then it's time to change to change your book. If it from some lecture note then you have to bear with it. Proving limits via $\epsilon, \delta$ is not an exercise in algebraic manipulation of inequalities. In general you can not use square roots here. $\endgroup$ – Paramanand Singh May 3 '17 at 5:37

I think you are asking about the choice of $\delta$. The idea is that if you choose $\delta = \min\{\delta_1,\delta_2\}$, then $|x-5|<\delta$ will imply both $|x-5|<\delta_1$ and $|x-5|<\delta_2$. In particular, it will imply both $-\delta_1<x-5$ and $x-5<\delta_2$.

Since you are using $\delta_1=5-\sqrt{25-\epsilon_2/3}$ and $\delta_2=-5+\sqrt{25+\epsilon_2/3}$, then you have the nicety that $|x-5|<\delta$ will imply $$-\underbrace{\left({5-\sqrt{25-\epsilon_2/3}}\right)}_{\delta_1}<x-5$$ and $$x-5<\underbrace{\left(-5+\sqrt{25+\epsilon_2/3}\right)}_{\delta_2}$$ so putting these together, you would get $$-\left({5-\sqrt{25-\epsilon_2/3}}\right)<x-5<\left(-5+\sqrt{25+\epsilon_2/3}\right)$$ i.e. $${-5+\sqrt{25-\epsilon_2/3}}<x-5<-5+\sqrt{25+\epsilon_2/3}$$ which is what you need to end up with the desired result, as you have shown.

As for the $\epsilon_2$, yes, this is fine. Since you need to show something is less than a given $\epsilon$, it suffices to show it is less than $\epsilon_2$ because you are creating $\epsilon_2$ which itself is less than $\epsilon$ by construction. Explicitly, if you show $|y-74|<\epsilon_2$ and you know $\epsilon_2<\epsilon$ by construction, then you have shown $|y-74|<\epsilon$.

I hope all of this answers your questions.

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  • $\begingroup$ Great answer, thanks. $\endgroup$ – Samson May 2 '17 at 14:29

To find values of $\epsilon$ and $\delta$. It is a good idea to do the following:

Let $\lim_{x\rightarrow c}f(x)=l$, to find $\delta$ which works for all epsilon in: $$\forall\epsilon>0\exists\delta>0:0<|x-c|<\delta\Rightarrow|f(x)-l|<\epsilon$$ Assume that $\epsilon>0$ is given. Now play with $|f(x)-l|<\epsilon$ to reach $|x-c|<F(\epsilon)$, where $F(\epsilon)$ is only a function of $\epsilon$.

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  • $\begingroup$ Noe that, some times you need to put some fixed upper or lower bound for some factors of $|f(x)-l|$ $\endgroup$ – SaeidAli May 2 '17 at 13:34
  • $\begingroup$ That is the method I usually use, I just wanted clarification on one of the steps in the above proof. Thanks. $\endgroup$ – Samson May 2 '17 at 13:38

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