TITLE: Understanding Epsilon-Delta Proof
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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 $1$, $\,\delta_f = \epsilon_f$.