I think you have a fairly decent idea, but the implementation is quite seriously flawed. As I pointed out in my comment, one of the mistakes in your work is that you cannot conclude that $|x^3 - 8| \lt \varepsilon$ given only $x^3 - 8 \lt \varepsilon$. (Exercise: Do you see why this is wrong? The trouble arises when $x$ is smaller than $2$, so that $x^3-8$ is a -- possibly large -- negative number.)
One way to remedy the proof is to fork into two cases: $x < 2$ and $x > 2$. That is, we define two different thresholds $\delta_+$ and $\delta_-$ that work separately for $x > 2$ and $x < 2$ respectively; then the overall $\delta$ is defined to be the smaller of the two.
Let's now see the above idea in action. Fix an $\varepsilon > 0$. Define $\delta_+ = (8 + \varepsilon)^{1/3} - 2$ and $\delta_- = 2 - (8 - \varepsilon)^{1/3}$; also define $\delta = \min \{ \delta_-, \delta_+ \}$. Note that $\delta_+, \delta_-$, and $\delta$ are all strictly positive; the proof would be incomplete without this observation. Now
when $2 < x < (2 + \delta_+)$, we have $0 < x^3 - 8 < \varepsilon$; and
when $(2 - \delta_-) < x < 2$, we have $- \varepsilon < x^3 - 8 < 0$.
Combining these two statements, we can write that whenever $2 - \delta_- < x < 2 + \delta_+$ and $x \ne 2$, we have $- \varepsilon < x^3 - 8 < + \varepsilon$. In particular, for $x \in (2 - \delta, 2 + \delta) \smallsetminus \{ 2 \}$, we have $|x^3 - 8| \lt \varepsilon$. We have thus showed that the limit of $f(x)$ as $x \to 2$ is $8$. $\qquad \diamond$
Although the above proof is correct, it is quite unsatisfactory because of many reasons. [This list is admittedly subjective and vague, so I recommend that you do not worry if something is unclear here.]
This style of argument relies in some sense on the fact that $f$ is monotonic. The monotonicity allowed us to “invert” the $\varepsilon$-$\delta$ condition in a straightforward way. (Moreover, it was useful that $f$ had a “nice” inverse.) For many functions, such a simple strategy does not work; so we often resort to establishing “bounds”. The textbook proof gives a good example of the latter approach.
The trick of considering the left and right sides separately works only in one dimension, i.e., the real line. It wouldn't work in more general spaces like $\mathbf R^2$, for instance. Once again, the textbook proof would generalise more easily.
In our proof, we were able to find a suitable $\delta$ without expending too much effort. In contrast, the textbook proof proceeds via a nontrivial estimate. Nevertheless, this effort does not totally go waste because the author manages to find a $\delta$ having a much simpler form; specifically, it is proportional to $\varepsilon$. This significance of this point will become more evident once you learn about derivatives, because the derivative of a function at a given point essentially tries to quantify the ratio $\varepsilon / \delta$ for small values of $\delta$.
Let us see the textbook proof now.
The textbook proof done “backwards”. For any polynomial $f$ and for any real number $a$, the difference $f(x) - f(a)$ is divisible by $x-a$. Therefore we can factor an $x-a$ out, and write $f(x) - f(a)$ as the product of $x-a$ and some other polynomial. Already this suggests that when $x-a$ is “small”, then the difference $f(x) - f(a)$ must also be small. However, to make this intuition precise, we proceed as follows.
In our example, $f(x) = x^3$ and $a=2$, so
$$
f(x) - 8 = (x-2) \cdot (x^2 + 2x + 4).
$$
As mentioned before, the $(x-2)$ factor is responsible for making the difference $(f(x) - 8)$ go to $0$ as $x \to 2$. On the other hand, the second factor $x^2 + 2x + 4$ approaches $2^2 + 2 \cdot 2 + 4 = 12$ as $x \to 2$. Inspired by this observation, we want to write that for $x$ close to $2$,
$$
f(x) - 8 \approx 12 (x-2). \tag{$\dagger$}
$$
Unfortunately, as intuitive as it might seem, this statement is neither precise nor correct, because we cannot selectively evaluate just one of the factors at the point $x=2$. Nevertheless this can be fixed because we only care about establishing an upper bound on the second factor when $x$ is close to $2$.
More precisely, for all $x \in (1, 3)$, we have
$$
|x^2 + 2x + 4| = x^2 + 2x+4 \leqslant 3^2 + 2 \cdot 3 + 4 = 19,
$$
which implies that
$$
|x^3 - 8| \leqslant 19|x-2| \tag{$\ddagger$}
$$
for all $x \in (1, 2)$. Comparing $(\dagger)$ and $(\ddagger)$, note that the right hand side slightly worsened from $12 |x-2|$ to $19 |x-2|$, but this is not of much consequence to us for the purposes of calculating the limit. All we want is some bound that goes to $0$, and $(\ddagger)$ works just fine.
[[EDIT: There is a close connection to derivatives here. Note that even though $(\dagger)$ doesn't make precise sense, the expression $12(x-2)$ feels like the “right” approximation to $(f(x)-8)$. In particular, the $19$ in $(\ddagger)$ is plainly arbitrary; we could have replaced it by any constant bigger than $12$ (for $x$ sufficiently close to $2$). In fact, we can think of $f(x) - 8$ as essentially $12(x-2)$, plus a “lower-order” correction term; derivatives formalise this idea nicely.]]
Finally, given $\varepsilon > 0$, we pick our $\delta$ such that both of the following conditions hold simultaneously:
First, for our bound $(\ddagger)$ to apply, we want our $x$ to lie in the interval $(1, 3)$, which requires $\delta$ to be smaller than $1$.
$(\ddagger)$ gives an upper bound of $19 |x-2|$ on $|f(x) - 8|$, so we want this upper bound to be at most $\varepsilon$. This forces the constraint $19\delta \leqslant \varepsilon$.
Of course, we could satisfy both these inequalities by picking $\delta = \min \{ 1 , \frac{\varepsilon}{19} \}$, which is exactly the choice made by the author. Now it is a matter of carefully doing the proof “forwards” to ensure that the whole argument works fine. I leave this as an exercise.
That seems a little bit confusing to me. For one thing there is a concern about the square roots being defined. I would recommend working as follows.
Note that $|x^2-4|=|x+2||x-2|$. Looking at this, we can make the factor of $|x-2|$ small, but we have to be sure that the factor of $|x+2|$ is not too big. But that's not so bad: if $|x-2|<1$ then $x \in (1,3)$, and so $|x+2|<5$. Hence $|x^2-4|<5|x-2|$. So if $|x-2|$ is also less than $\frac{\varepsilon}{5}$ then $|x^2-4|<\varepsilon$.
So now you're done: given $\varepsilon > 0$, choose $\delta=\min \{ 1,\varepsilon/5 \}$. With this choice, if $|x-2|<\delta$ then $|x^2-4|<\varepsilon$.
This kind of approach will work in a lot more problems than your kind of approach, because usually you can't find the optimal bounds like you tried to do. Instead you usually make a series of estimates, which are hopefully accurate enough to get you what you need.
Best Answer
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".