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,
$$0 < |x-4| < \delta = \min \{ 1, \frac{\epsilon}{27} \}$$
 A: Example of proof-writing:  Let  $\epsilon >0.$ Let $\delta =\min (1,\epsilon /27).$ $$\text { Then }\;\; \delta >0, \;\;\text  { and } \; 0<|x-4|<\delta \implies |3x^2-48|<\epsilon.$$ Proof: (i). That $\delta >0:$ Obvious.
$$\text {(ii). }\;  0<|x-4|<\delta \implies  |3x^2-48|=3\cdot |x-4|\cdot |x+4|=$$      $$=3\cdot |x-4|\cdot |(x-4)+8|\leq$$    $$\leq 3\cdot |x-4|\cdot (|x-4|+8)\leq$$ $$\leq 3\cdot |x-4| \cdot  (\delta +8)\leq $$ $$\leq 3\cdot |x-4|\cdot 9=27 |x-4|<$$ $$<27 \delta \leq 27 (\epsilon /27)=\epsilon.$$
In your work in your edited version, you did some successful "exploratory surgery" to find that  $\delta =\min (1.\epsilon /27)$ would suffice. Essentially the implications in your work are going from right to left, so when you reach $\delta =\min (1,\epsilon /27)$ you should then present a re-arrangement   into the form I have shown, so that you present  a clear and complete picture of the "logical flow".... Or,if you prefer, just present the finished proof  with no explanation of how you found it, in the style of Carl F. Gauss.
A: Analisys is the art of bounding expressions.
In this case we have to bound $|3x^2 - 48|$ by $\epsilon$ when $x$ is not too far from $4$. Here is how
$$
|3x^2 - 48| = 3|x^2 - 16| = 3|x - 4||x + 4| \le 3|x - 4|c
$$
provided that $|x + 4| \le c$ if $x$ is close enough to $4$.
How close? That's up to us. For instance, we can impose $3 \le x \le 5$, which means $|x - 4| \le 1$.
So, $7 \le x + 4 \le 9$ and therefore $|x + 4| = x + 4 \le 9$. Hence, we can take $c = 9$. Replacing
$$
|3x^2 - 48| \le 3\cdot9|x - 4|
$$
and if we put $3\cdot9|x - 4| < \epsilon$ we get $|x - 4| <\epsilon/27$.
In sum, we have two restrictions for $|x - 4|$. Firstly, it must be $\le1$ and secondly $<\epsilon/27$. In other words, we have proven that
$$
|x - 4| < \min(1, \epsilon/27) \implies |3x^2 - 48| < \epsilon
$$
A: Here comes another discussion (and limit proof):
Let $\epsilon>0$ be given. We want to find $\delta>0$ such that $|3x^2-48|<\epsilon$ whenever $|x-4|<\delta$.
We have $|3x^2-48| = 3 |x+4| |x-4| < 3 |x+4| \delta$ when $|x-4|<\delta$. If we were to set $\delta = \epsilon/(3 |x+4|)$ then we get $|3x^2-48| < \epsilon$, but we can not do this since $\delta$ may not depend on $x.$ 
The factor $|x+4|$ thus is problematic. But using the triangle inequality we can eliminate it: $|x+4| = |(x-4)+8| \leq |x-4| + |8| < \delta + 8.$ Thus we have $|3x^2-48| < 3 (\delta+8) \delta.$
If we set $\delta$ so that $3(\delta+8)\delta = \epsilon$, i.e. $\delta = -4 + \sqrt{16 + \epsilon/3}$ (note: taking a minus sign makes $\delta<0$) we get $|3x^2-48| < \epsilon$. We can do this, but it's ugly having to solve a quadric, and in other cases we can not solve for $\delta$. We therefore want to eliminate the factor $\delta+8$ and only keep the linear factors.
By always making sure that $\delta \leq 1$ we get $\delta+8 \leq 9$ so that we have $|3x^2-48| < 27 \delta.$ We now can take $\delta = \epsilon/27$ or at least $\delta \leq \epsilon/27.$
Thus we take $\delta \leq 1$ and $\delta \leq \epsilon/27,$ i.e. $\delta \leq \min(1, \epsilon/27),$ and then we have $|x+4| \leq 9$ so that
$$|3x^2-48| = 3 |x+4| |x-4| < 3 \cdot 9 \cdot \delta = 27 \delta \leq \epsilon$$
whenever $|x-4| < \delta.$
A: 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.
