Is my understanding of the definition of limit correct? The definition;

$(\forall \epsilon > 0)(\exists \delta > 0)(\forall x$ satisfying $0 <$
  $|x-a| < \delta$ also satisfies $|f(x) - L| < \epsilon) \iff \lim_{x\to a} f(x) = L$ 

To explain my understanding, lets consider the following example;

$$\lim_{x\to 4} 3x^2 = 48.$$

Proof:
Let $\epsilon > 0$ be given such that $|3x^2 - 48| < \epsilon.$Then observe that 
$$|3x^2 - 48| = 3 \cdot |x-4| \cdot |x+4| < \epsilon.$$
Now, since our aim is to find a $\delta > 0$(probably as a function of $\epsilon$) such that for all $x$ satisfying $0 < |x-4| < \delta$, we have $|3x^2 - 48| < \epsilon$, i.e whatever the value of $\epsilon$ is, we just want to make sure that the condition is satisfied, so to be, sort of, safe and get rid of the factors, we can restrict ourselves by assuming $|x-4| < 1$, so this implies $|x+4| < 9$, so we have 
$$3 \cdot |x-4| \cdot |x+4| < 27 |x-4|,$$
and we already assumed that 
$$3 \cdot |x-4| \cdot |x+4| < \epsilon.$$
Now, if $\epsilon < 27 |x-4|$, there will be some $x$ such that 
$$3 \cdot |x-4| \cdot |x+4| \not < \epsilon,$$
which would contradict with out assumption, so we must have 
$$27 |x-4| \leq \epsilon,$$
which implies $|x-4| \leq \frac{\epsilon}{27}.$
However, this result is only valid if $|x-4| < 1$, so we must state that
$$0 < |x-4| < \min\left\{1, \frac{\epsilon}{27}\right\} = \delta$$
QED.

First of all, is there any flow or misunderstanding in my understanding of the definition of the limit and generally with the proof of the example.
Secondly, in the book that I'm using it says that we don't want to make $|x-4|\cdot |x+4|$ too large, why ? what is wrong with it ? what if we do, then what ?
Edit:
As @md2perpe pointed out in his/her comment, there is logical mistake in the proof, so I'm writing the, so called, corrected version;
Let $\epsilon > 0$ be given. 
$$|3x^2 - 48| < \epsilon \iff 3 \cdot |x-4| \cdot |x+4| < \epsilon,$$
so let assume $|x-4| < 1$, hence $|x+4| < 9$, so we have 
$$|3x^2 - 48| = 3 \cdot |x-4| \cdot |x+4| < 27 |x-4|.$$
Since we want to find $\delta(\epsilon) > 0$, we want to relate $\epsilon$ to $\delta$, somehow, so if we say $\epsilon < 27 |x-4| < 27 \delta$, there will be some $x$ such that $0 < |x-4| < \delta$ but $|3x^2 - 48| \not < \epsilon$, so we consider the case $27 |x-4| < \epsilon$.Thus,
$$0 < |x-4| < \delta = \min \{ 1, \frac{\epsilon}{27} \}$$

I'll try to answer your questions one at a time:

First of all, is there any flow or misunderstanding in my understanding of the definition of the limit?

No, you have stated the definition of the limit accurately. I would have dispensed with the logical symbols and simply said that $\lim_{x\to a}f(x) = L$ if corresponding to each $\epsilon>0$ is a $\delta>0$ such that $|f(x)-L| < \epsilon$ whenever $0<|x-a|<\delta$, however.

Is there any flow or misunderstanding in my understanding of the proof of the example?

Yes, there is at least one point where you are tripped up. You have the following:

Since we want to find $\delta(\epsilon) > 0$, we want to relate $\epsilon$ to $\delta$, somehow, so if we say $\epsilon < 27 |x-4| < 27 \delta$, there will be some $x$ such that $0 < |x-4| < \delta$ but $|3x^2 - 48| \not < \epsilon$, so we consider the case $27 |x-4| < \epsilon$.

At this point in your proof, you are already done. This entire paragraph can be eliminated in favor of saying
If $|x-4| < 1$, then $|x+4|<9$, hence $|3x^2 - 48| = 3 |x-4| \cdot |x+4| < 27 |x-4|$. Consequently, if we set $\delta=\min(1,\epsilon/27)$, the claim follows. $\qquad \square$
It's not just about style. When you are considering separately the cases $\epsilon>\text{something}$ and $\epsilon<\text{something}$, you are opening up opportunities for innacuracy. In each $\epsilon$-$\delta$ proof, we never have to make assumptions on $\epsilon$ since $\epsilon$ is handed to us, and we have to work around it.

Secondly, in the book that I'm using it says that we don't want to make $|x−4|\cdot|x+4|$
  too large. Why? What is wrong with it? If we do, then what?

The idea is that we want the quantity $|x−4|\cdot|x+4|$ to be small, and to do this, we really want to get some control on the quantity $|x+4|$, since if we had a bound like $|x+4|<M$ for some $M$, we could do something like set $\delta < \epsilon/M$ to finish up.
