Use the $\epsilon-\delta$ definition of limit to prove that $\lim\limits_{x\to2}x^2 = 4$ I am studying Calculus myself and confused by this example. 

EXAMPLE: Use the  $\epsilon-\delta$ definition of limit to prove that $$\lim\limits_{x\to2}x^2 = 4$$
  Solution: You must show that for each $\epsilon>0$, there exists a $\delta$>0 such that 
  $$|x^2-4|<\epsilon$$whenever
  $$0<|x-2|<\delta$$
  To find an appropriate $\delta$, begin by writing $|x^2-4|=|x-2||x+2|$. For all $x$ in the interval (1,3), $x$+2<5 and thus $|x+2|<5$. So, letting $\delta$ be the minimum of $\epsilon/5$ and 1, it follows that whenever $0<|x-2|<\delta$, you have$$|x^2-4|=|x-2||x+2|<\frac{\epsilon}{5}5=\epsilon$$
  For $x$-value within $\delta$ of 2(x not equal 2), the values of $f(x)$ are within $\epsilon$ of 4.

Question 1: Can I consider for all $x$ in the interval(-12, 8), instead of (1,3)? In that case, $|x+2|<10$
Question 2: In the solution, it says "letting $\delta$ be the minimum of $\epsilon/5$ and 1". Can I just say that "letting $\delta$ equal $\epsilon/5$"? What is the point of mentioning "1" here?
Question 3: What is the point of say "For $x$-value within $\delta$ of 2(x not equal 2), the values of $f(x)$ are within $\epsilon$ of 4." at the end of the solution? I don't see any value of this sentence.
 A: *

*No, it has to be in the form of $(2-\delta_0,2+\delta_0)$(note that we don't look at $2$ itself), note that then the value of $\delta$ will change

*They choose that $x$ is in $(2-1,2+1)=(1,3)$, so we put a restriction on $\delta$, so before letting $\delta=\epsilon/5$ we need to make sure we are at least in the restriction, hence the minimum

*This part is just to explain what we got: when the distance of $x$ from $2$ is at most $\delta$ the distance of $f(x)$ from $4$ is at most $\epsilon$ 

A: Your questions are about the routine but at first seemingly random ideas that you need to turn understanding into a formal proof. The reason we write
$$
|x^2 - 4| = |x-2||x+2|
$$
is that we know we have to control the size of the left member by restricting the difference between $x$ and $2$.
So we say to ourselves "how can we bound the other factor on the right?" Well, any restriction on $x$ will do that. It's traditional to use the interval $(1,5)$ but any other interval will do. You suggested $(-12, 8)$, which would be fine, but unconventional. Then the $1/5$ in the proof would become $1/14$.
In order to use that $1/5$ (or $1/10$ you need to be sure your bound on the distance between $x$ and $2$ puts $x$ in the interval. That requires that $\delta$ be less than $1$ in the argument you quote. If you used the interval $(-12, 8)$ you's specify $\delta < 6$ as well as less than $\epsilon/14$.
The answer to your third question is that the last sentence is just an elaborate way to say "QED". It restates the fact that you proved what you needed to prove to show the limit is what you say it is.
Note: @Holo 's answer is essentially the same as mine. This one is a little wordier, in hopes of helping you distinguish between the parts of the proof that require understanding, and the parts required to make it airtight.
