Question about the inequality in the proof of $\lim_{x\to 4}{\sqrt{x}}=2$ by $\epsilon$ - $\delta$ method? What I want to proof is
$\lim_{x\to 4}{\sqrt{x}}=2$
The following is from my textbook
Finding a $\delta$. Let $\epsilon\gt 0$. We seek a $\delta\gt 0$, s.t.
$if\quad 0\lt|x-4|\lt\delta,\quad then\quad|\sqrt{x}-2|\lt\epsilon$
To be able to form $\sqrt{x}$, we need to have $x\ge 0$. To ensure this, we must have $\delta\le 4$.
How to derive/translate this statement in inequality?
 A: Note first that $|x-4|=|\sqrt x-2||\sqrt x+2|$ when $\sqrt x$ exists.
Now, let $\delta=\min(4,2\epsilon)$. Then if $|x-4|<\delta$, $\sqrt{x}$ exists (in particular, $|\sqrt x+2|\ge 2$), and
$$|\sqrt x-2|=\frac{|x-4|}{|\sqrt x+2|}\le\frac{1}{2}|x-4|<\frac{1}{2}(2\epsilon)=\epsilon.$$
Look at what has happened here. Why did we choose $\delta=\min(4,2\epsilon)$? We needed $\delta\le 4$ because this ensures that $\sqrt x$ exists, and we needed $\delta\le 2\epsilon$ to ensure that the inequality above worked out. Therefore the minimum of the two sufficed.
Edit: I will show that if $\delta\le 4$, then $|x-4|<\delta$ implies $x\ge0$. 
Let $|x-4|<\delta$. If $x>4$, then of course $x>0$. Otherwise, $x\le 4$ and we have
$$4\ge\delta>|x-4|=4-x$$
which implies $0\ge -x$, i.e. $x\ge0$.
I think the way you should think of this intuitively is in terms of distances. That is, you should read $|x-4|<\delta$ as "the distance between $x$ and $4$ is smaller than $\delta$".
In particular, if $\delta\le 4$ then $|x-4|<\delta$ says "the distance between $x$ and $4$ is less than $4$", which means $x$ must be greater than $0$.
