Clarification of epsilon-delta limit proof

Question

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

$0\lt|x-5|\lt\delta$

Implies,

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

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

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.

Answer 1

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.

Answer 2

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 $\epsilon$.