Understanding Epsilon-Delta Proof I'm trying to understand the proof of the following limit using epsilon-delta definition.
$$\lim _{x\to3} x^2 = 9$$
In Stewart Calculus, the proof goes like this:

I'm confused at the following points:
1) How the author arrives at this result: $|x-3| < \varepsilon/C = \delta$? 
Specifically, how,  $$|(x+3)(x-3)| < \varepsilon$$ and $$|(x+3)(x-3)| < C|(x-3)|$$ leads to $$|x-3| < \varepsilon/C = \delta.$$ This is the part I've trouble understanding.
2) What's the role played by $\delta=\min(1, \varepsilon/7)$ ?
I am trying to understand this epsilon-delta proofs for a week but to no avail.
 A: I think your main issue is that you are still trying to think about this exercise as a routine algebraic manipulation. It is not exactly like that.
The thing is that here we have a goal/target about ensuring that some inequality holds. In current question the goal is to ensure that $$|x^2-9|<\epsilon$$ We are not supposed to find all values of $x$ for which the above inequality holds (similar to solving equations like $x^2=9$). The problem is not exactly algebraic. Rather what we desire is to find a range of values of $x$ near $3$ for which this inequality can be ensured. Such a range of values of $x$ may or may not exist. Our task is to prove that such a range of values of $x$ near $3$ always exists no matter what $\epsilon $ is given.
The technique is to replace the target inequality by a simpler one. Thus we have to find some expression $g(x) $ which is simpler in form and satisfies $$|x^2-9|<g(x)$$ and then replace the goal with ensuring that $g(x) <\epsilon $. Thus our original target is to be achieved via a combination of two simpler goals $|x^2-9|<g(x)$ and $g(x) <\epsilon$.
The problem is now to choose a suitable $g(x) $ and to find a range of values of $x$ near $3$ which can ensure that both the sub-goals are met. This is where one has great leverage and problem is actually far simpler than it appears. We have $$|x^2-9|=|x+3||x-3|$$ Now let us choose any specific range of values of $x$ near $3$, say $|x-3|<1$ (this is totally as per your wish, but in general the range should be such that the desired simplification in what follows is possible). And $$|x+3|\leq |x-3|+6<7$$ and therefore we have $$|x^2-9|=|x+3||x-3|<7|x-3|$$ for the range of values of $x$ given by $|x-3|<1$.
Thus we can choose $g(x) =7|x-3|$ and one of the sub goals is achieved for the range $|x-3|<1$. The other goal is now simpler $$7|x-3|<\epsilon $$ Obviously this can be achieved by the range of values of $x$ given by $|x-3|<\epsilon /7$ (if this is not obvious to you then you need to see how inequalities work in general).
So for the two goals we have found two ranges of values of $x$ namely $|x-3|<1$ and $|x-3|<\epsilon /7$ which ensures that the respective goals are met. Since we want to ensure that both the goals are met simultaneously we need to deal with the range of values of $x$ which are common to both $|x-3|<1$ and $|x-3|<\epsilon/7$. This is possible if $|x-3|<\min(1,\epsilon /7)$ and we are done by setting $\delta=\min(1,\epsilon/7)$ and our desired range of values of $x$ is $|x-3|<\delta$.

The important thing to notice here is that our original problem to ensure some inequality is replaced by two far simpler (but not necessarily equivalent) problems. This is in quite contrast to solving equations like $x^2-9=0$ where the problem is reduced to two simpler and equivalent problems $x-3=0,x+3=0$.
The fact that we have to simplify the problem without caring about equivalence gives us great leverage here. Most beginners however don't notice this and instead focus on solving inequalities (where the problem can be simplified but only to an equivalent one) and this is one of stumbling blocks in understanding and applying definition of limit.
More formally the target inequality $$|f(x) - L|<\epsilon $$ is not a hypothesis but a conclusion in a long chain of logical implications. Also by definition the implications involved are one way and you don't need to put any extra effort to unnecessarily ensure a both way implication. And we present our argument like "the target conclusion, say $A$, holds if (not iff) $B, C, \dots$ hold and so on till we reach a stage where we get to see ranges of values of $x$". So the chain of implications is figured out in reverse.
Using your own words from question: how $$|x+3||x-3|<\epsilon$$ and $$|x+3||x-3|<C|x-3|$$ lead to $$|x-3|<\epsilon /C$$ is not the right question, but you should ask how $$|x-3|<\epsilon /C$$ and $$|x+3||x-3|<C|x-3|$$ lead to $$|x+3||x-3|<\epsilon $$ This is the desired logical flow and it would now look obvious to you. The thing however is that the individual logical implications have to be figured out in reverse starting from the conclusion to the hypotheses.
Years of training in algebraic manipulation which are mostly forward or both way implications makes things in analysis a little bit surprising (if not difficult) when we have to deal with one way implications in reverse manner. Thus we switch from "$A$ implies $B$" to "$B$ holds if $A$ holds". 
A: *

*The author is just saying that if $|x+3||x-3|<C|x-3|$, then$$|x-3|<\frac\varepsilon C\implies|x+3||x-3|<C|x-3|<C\frac\varepsilon C=\varepsilon.$$

*If you define $\delta=\min\left\{1,\frac\varepsilon7\right\}$, then, if $|x-3|<\delta$, you know that $|x-3|<1$ and that $x-3|<\frac\varepsilon7$. And the author proved that, when both of these inequalities hold, $|x+3||x-3|<\varepsilon$.

A: Proposition 1: Let $p, v, M \in \Bbb R$ with $v,M \gt 0$.
Let $f$ be a real-valued function that is defined over the interval $[p-v,p+v]$ satisfying
$\tag 1 \displaystyle {\lim _{x \to p} f(x) = 0}$
Let $g:[p-v,p+v] \to \Bbb R$ be given and satisfy $|g(x)| \le M$ for all $x$ in its domain.
Then the function $h(x) = f(x)g(x)$ defined over $[p-v,p+v]$ satisfies
$\tag 2 \displaystyle {\lim _{x \to p} h(x) = 0}$
Proof
Let the challenge $\varepsilon \gt 0$ be submitted for $\text{(2)}$.
By $\text{(1)}$, for the number $\varepsilon_f = \frac{\varepsilon}{M}$  a corresponding $\delta_f \gt 0$ can be specified such that
$\quad 0 \lt |x - p| \lt \delta_f \text{ implies } |f(x)| \lt \varepsilon_f$
Set $\delta = \text{min}(\delta_f,v)$. It is easy to see that 
$\quad 0 \lt |x - p| \lt \delta \text{ implies } |h(x)| \lt \varepsilon$
and so the validity of $\text{(2)}$ has been established. $\quad \blacksquare$
For the OP's problem, set
$f(x) = x - 3$
$g(x) = x + 3$
$p = 3$
$v = 1$
Now if we check we see that the $|g(x)| \le 7$ on $[2,4]$ so set $M = 7$.
For any $\varepsilon$ presented for $\displaystyle \lim_{x\to3} x^2 = 9$,  we can '$\text{turn  the proposition 1 crank}$', and let 
$\quad \delta = \text{min}(\frac{\varepsilon}{7},1)$
Note that since $f(x) = x -1$ has slope equal to $1$, $\,\delta_f = \epsilon_f$.
A: It's a puzzle you are working backwards to find  values that work.
We have $\epsilon>0$  we found it on the ground and said "Hey, lets use this epsilon".
We are imagining and daydreaming about wouldn't it be wonderful if we had a $C$ so that $|x+3| < C$.  We don't actually have this but we are imagining "what if".
If we had such a $C$ it would be a positive number.  And $\epsilon$ is a positive number.  So $\frac {\epsilon}C$ is a positive number.
So we had that dreamy magical wouldn't it be nice $C$ we could just take $\frac \epsilon C$ and call it $\delta$.  Let's imagine we did that.
Then we can pick an $x$ so that  $|x-3| < \delta$.  We can do that by noting that $\delta$ is a positive number, so we just pick some $x$ that is within $\delta$ of $3$.  That is picking $x$ so that $3-\delta < x < 3+\delta$.
Okay we did that.
So $|x-3| < \delta = \frac \epsilon C$.  That fine.  
Now.... just multiply both sides by $C$.  ... You can.... $C$ is a positive number so just multiply both sides by it. 
$C|x-3| < \delta*C = \frac \epsilon C*C = \epsilon$.
Well, that's handy.
Now $C > |x+3|$.  Why?  Well, because we really want it to be. Remember we don't actually have $C$ yet.  We are only daydreaming (well, speculating) what would have if we had a $C$ so that $C > |x+3|$.
Well, if we had and $|x+3 | < C$ we could multiply both side but $|x-3|$ and get
$|x-3||x+3| < C|x-3|$ and we have $C|x-3| < \epsilon$ so we would have
$|x-3||x+3| < C|x-3| < \epsilon$.  
And that would be freaking awesome! Because we would be done and have proven every thing we want to prove.
We can pick any $\epsilon > 0$ of the ground.  We take the $C$ we have.  We let $\delta = \frac {\epsilon}C$.  Then whenever we pick $x$ so that $|x-3| < \delta$ we have $|x^2 -9| = |x-3||x+3| < |x-3|C < \delta C = \epsilon$ and that would prove $\lim_{x\to 3} x^2 = 9$.
....
But... we have to have that $C$. And so for $C$ is just a daydream.
But what do we need to find the $C$?  We need $|x-3| < \delta$ and we need $|x+3| < C$.  We want both of those to be true. 
Now we want $\delta$ to be small so we can assume $\delta < 1$ and $x$ is within $1$ of the number $3$.  We don't actually know what $\delta$ will until we figure out what $C$ is but we don't know what $C$ is.  So we fudge.  We'll assume $\delta < 1$.  If by some bad bit of luck we get $\delta = \frac {\epsilon} C \ge 1$ well, well ust pick $\delta = 1$ instead.
So $x$ is within $\delta$ of $3$ then $2 < x < 4$ and we so $5 < x+3 < 7$ and $|x+3| < 7$.
So that is our magincal daydream number Let $x = 7$. And let $\delta =\frac \epsilon 7$.  That will work if $\frac \epsilon 7 \le  1$.
But if $\frac \epsilon 7> 1$ we can use $\delta = 1$.
.......
That's how we puzzled it all out.... but lets see if it works:
We pick up an $epsilon $ off the ground.  
We let $\delta = \min (\frac \epsilon 7, 1)$ that is if $\epsilon \ge 7$ we let $\delta =1$ and if $\epsilon < 7$ we let $\delta = \frac \epsilon 7$.
We pick an $x$ so that $3-\delta < x < 3+\delta$ or,  in other word, so that  $|x-3| < \delta$.
If $\epsilon \ge 7$ and $\delta =1$ then $2 < x < 4$ and $4 < x^2 < 16$ and  so $-5 < x^2 - 9 < 7$ and $|x^2 - 9| < 7 \le \epsilon$ and we are done.  We found there is a $\delta$ for all $\epsilon \ge 7$  that satisfies what we want.
But that was the dumb case.  We have to also show it for all the $\epsilon < 7$.
Okay.  If $|x-3| < \delta = \frac {\epsilon}7\le 1$ that's a start.
But then $2< 3-\delta < x < 3+\delta < 4$ so $5 < x +3 < 7$ so $|x+3| < 7$
So $|x^2-9| = |x-3||x+3| < \delta*7 = \frac {\epsilon}7*7 = \epsilon$.
And that's it!  We proved it.  For every $\epsilon > 0$ there is a $\delta$ so that $|x-3| \implies |x^2 -9| < \epsilon$.  We just had to make for that if $\epsilon \le 7$ that $\delta = \frac \epsilon 7$ and if $\epsilon > 7$ that $\delta = 1$.
(Actually we could pick an delta smaller if we wanted to.)
