Here is the example given in the book (which I am paraphrasing):
We have the function $f(x)=x^2$. We should be able to show that $f(x)$ approaches $9$ as $x$ approaches $3$; in other words, that $|x^2-9|<\epsilon$ for an arbitrary positive integer $\epsilon$ if we get $|x-3|$ small enough. So we want $|x-3|$ such that $$|x^2-9|=|x-3| \dot |x+3|<\epsilon.$$
But this is not so easy because of the term $|x+3|$, so it would be nice to find a way to write $|x+3|$ in terms of some constant, making the inequality easier to manipulate. So we choose a restriction on $|x-3|$ in order to help us determine the value of $|x+3|$. Specifically, we say that $|x-3|<1$, or $2<x<4$, so $5<x+3<7$ and we've guaranteed that $|x+3|<7$. So then $$|x^2-9|=|x-3| \dot |x+3|<7|x-3|$$ and therefore $|x^2-9|<\epsilon$ if $|x-3|<\frac \epsilon 7$ and $|x-3|<1$, which is the same as saying $|x^2-9|<\epsilon$ if $|x-3|<$min$(\frac \epsilon 7,1)$.
What I don't understand: (1) Why are we allowed to choose a restriction on $|x-3|$? Also, (2) when other specifications are chosen, the inequality changes [for example, if we choose $|x-3|<10000$ instead of $|x-3|<1$, we get $|x-3|<$ min$(\frac \epsilon {10006}, 10000)$ instead of $|x-3|<$min$( \frac \epsilon 7, 1)$. And min$(\frac \epsilon {10006}, 10000)$ is actually less than min$( \frac \epsilon 7, 1)$. It seems that as our choice of $n$ for $|x-3|<n$ gets larger, the constraint $\frac \epsilon {n+6}$ grows smaller, which seems mysterious to me even though it makes sense algebraically. Wouldn't this mean that as we choose a wider initial interval of what $|x-3|$ can be, $|x-3|$ grows smaller in relation to $\epsilon$? Why is this? What do these differences actually mean? And also, how can the choice be arbitrary, if we keep getting different results?
What we want to show is this:
"For each $\epsilon>0$, there exists $\delta$ such that $|x-3|<\delta$ implies $|x^2-9|<\epsilon$".
So to prove it, let's fix an arbitrary $\epsilon>0$.
Then, we need to find $\delta$ which restricts $x$ to satisfy the above sentence.
Moreover, note that if you find such $\delta$, smaller $\delta' <\delta$ would make the sentence true too.
So you try one $\delta$, and if that doesn't work well, you can try smaller $\delta' < \delta$ then.
With this in mind, let's try your example.
First, simply take an arbitrary $\delta>0$.
Now, choose $|x-3|<\delta$.
Then, $|x^2-9|=|x-3||x+3| \leq \delta |x+3|$.
Now, we need to show that $\delta |x+3|<\epsilon$ to show $|x^2-9|<\epsilon$, but since we don't know how small $\delta$ is, we cannot bound $|x+3|$.
So let's try $\delta=1$.
Then for each $x$ satisfying $|x-3|<\delta$, $|x^2 -9 |=|x-3||x+3| \leq 4|x-3|$.
However, since we don't know how small $x$ and $\epsilon$ are, we cannot bound $4|x-3|$ smaller than $\epsilon$.
So let's try new $\delta=min\{ 1, \epsilon /4\}$.
Then, for each $x$ satisfying $|x-3|<\delta$, $|x^2-9|=|x-3||x+3|\leq 4|x-3| < 4\delta \leq \epsilon$.
So this $\delta$ works.
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You must get used to this "trying $\delta$" algorithm so that you find a sufficient $\delta$ within little tries.