Hint: First, note that $\lfloor x \rfloor$ is always an integer. Use this fact along with the fact that $x -1 < \lfloor x \rfloor \leq x$.
EDIT: By the way, your goal is to show that for all $\epsilon > 0$, there exists a $\delta > 0$, such that
$$0 < |x - 3 | < \delta \Longrightarrow \left| \left\lfloor \frac{x}{2} \right\rfloor - 3\right| < \epsilon.$$
This should be doable with the two previous observations. Your observation you mention is correct, but is not helpful (for your proof, anyways) because your interval $(1/2, 5/2)$ includes two integer values. On the other hand, the inequalities $x - 1 < \lfloor x \rfloor \leq x$ show that $\lfloor x \rfloor$ is the unique integer in the interval $(x-1, x]$.
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.
Best Answer
Hint:
$$\frac{x^2+x-6}{x^2-9}=\frac{(x+3)(x-2)}{(x+3)(x-3)}=\frac{x-2}{x-3}\,\,,\,\,x\neq \pm\,3$$