Understanding the limit definition in Stewart's Calculus I'm currently reading Stewart's Calculus and I'm stuck on "The Precise definition of limits" topic.
The author is solving one of the examples of finding (I don't know how to call them properly) "boundaries" of limits.
He gives an example as follows:

if $$0 < |x-3| < Q$$ then $$|x^2 - 9| < W$$
and he is solving those inequalities to find $Q$ in terms of $W$.

First he transforms $$|x^2 - 9| < W$$
to $$|x - 3||x + 3| < W$$
and here foes the part which I don't really understand:
He says that

Notice that if we can find a positive constant C such that $|x + 3| < C$, then
$$|x - 3||x + 3| < C|x-3|$$
(above part is clear for me)

and here we go:

"and we make $C |x-3| < W$ by taking $|x - 3| < W/C$ so we could choose $Q = W/C$"

I don't understand where $$C|x-3|<W $$
comes from. Can someone explain it to me, please?
 A: We can choose $x$ arbitrarly get closer to $3$ such that $|x-3|<\frac W C \iff C|x-3|<W$ and then
$$|x^2 - 9|=|x - 3||x + 3|<C|x-3|<W$$
A: He has shown so far that if you can find positive $C$ such that $\lvert x+3\rvert < C,$
then
$$\lvert x^2 - 9\rvert = \lvert x-3\rvert \lvert x+3\rvert < C\lvert x-3\rvert. $$
Or more simply $\lvert x^2 - 9\rvert < C\lvert x-3\rvert. $
Now you can choose any positive value of $Q$ that suits your purpose
in a problem like this.
So let's choose $Q = \frac WC$. Then if $\lvert x-3\rvert < Q$,
$$ C\lvert x-3\rvert < C\cdot Q = C \cdot \frac WC = W. $$
Putting this together with $\lvert x^2 - 9\rvert < C\lvert x-3\rvert$
you now have $\lvert x^2 - 9\rvert < W,$ as required.
The trick now is to make sure you have a value of $C$ that actually makes all of this work.
The proof is not complete until it actually shows what the value of $C$ should be.
I don't have a copy of Stewart's Calculus at hand,
so I don't know exactly how he wraps up this proof.
Usually in a proof like this we wouldn't need to use the symbol $C$ to represent some yet-to-be-determined number;
we would just write an explicit number, for example, $7,$
having previously figured out how we could makes sure that it was always true that $\lvert x + 3\rvert < 7.$
But I think what he's trying to do here is to lead you through the process of how you figure out the pieces that need to go into a proof like this,
as opposed to simply writing out the proof after you've figured out the pieces.
Note that if $Q$ is a very small number, then $\lvert x - 3\rvert$
will be an even smaller number,
so it's not necessarily a problem if $C$ turns out to be much larger than $W.$
For example, if $C = 7$ and $W = 0.01,$ but $Q = 0.001,$ then
$\lvert x - 3\rvert < 0.001$ and therefore
$C \lvert x - 3\rvert < 7\times 0.001 = 0.007,$
so it is certainly true that $C \lvert x - 3\rvert < 0.01.$
If $Q = \frac WC,$ as suggested, then the smaller $W$ gets the smaller $Q$ gets, and the larger $C$ gets the smaller $Q$ gets again, so there is a mechanism already built into this passage in the textbook to ensure that $Q$ is small enough to force $C \lvert x - 3\rvert$ to be less than $W.$
Again, I don't know exactly where Stewart goes with this next,
but I'm fairly certain that if you continue to read on,
there will be further explanation of what he did here.
Maybe he's next going to show how to decide exactly what value $C$ should actually have, or maybe he never actually finishes this proof but starts a new one where he doesn't need some fictional number that he just supposes might exist.
As you get into calculus and higher math you can often run into things in textbooks that don't make sense in isolation. Sometimes you have to go back in the book a bit to pick up the context that would explain what you're looking at, but sometimes you have to read ahead a bit to see where it goes next.
