Let $x \ge 0$. Determine a condtion on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$ I'm trying to understand the logic of this proof.
Let $x \ge 0$. Determine a condition on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$
Proof 
$|\sqrt{x} - 2| = \frac{|(\sqrt{x} - 2)(\sqrt{x} + 2)|}{\sqrt{x} + 2} = \frac{x-4}{\sqrt{x} + 2} \le \frac{|x-4|}{2}$
For $x$ satisfying $|x-4| = 2 \cdot 10^{-2}$
we have $|\sqrt{2} -2| \le \frac{2 \cdot 10^{-2}}{2}=10^{-2}$
Does the $|\sqrt{2} -2|$ just come from the 2 in $|x-4| = 2 \cdot 10^{-2}$? Is there more too it? Or is there a better, more complete proof? 
Thank you, in advance. 
 A: When you say $$|\sqrt{x} - 2| = \frac{|(\sqrt{x} - 2)(\sqrt{x} + 2)|}{\sqrt{x} + 2} = \frac{|x-4|}{\sqrt{x} + 2} \le \frac{|x-4|}{2}$$ The final $\le$ comes from replacing $\sqrt x +2$ with $2$.  As $\sqrt x \ge 0$, this is decreasing the denominator, which increases the fraction. The overall inequality
$$|\sqrt{x} - 2|\le \frac{|x-4|}{2}$$ is then used. We want to require that $|\sqrt x - 2| \lt 0.02$, so we demand that $$\frac{|x-4|}{2}\lt 0.02$$, which gives $$|x-4| \lt 0.01$$
A: What you need to do is find bounds for $\frac{1}{\sqrt x + 2}$ 
Suppose $|x - 4| < \delta$. Then we can argue that
\begin{align}
    |x - 4| < \delta
    &\implies4-\delta < x < 4 + \delta \\
    &\implies \sqrt{4-\delta} < \sqrt x < \sqrt{4 + \delta}\\
    &\implies \sqrt{4-\delta} + 2 < \sqrt x + 2 < \sqrt{4 + \delta} + 2\\
    \text{If we also require that }\; &0 < \delta < 3 \;
    \text{then we can continue with}\\
    &\implies 3 < \sqrt x + 2 \\
    &\implies \dfrac{1}{\sqrt x + 2}< \dfrac 13 \\
    &\implies |\sqrt x - 2| < \dfrac{|x - 4|}{\sqrt x + 2} \\
    &\implies |\sqrt x - 2| < \dfrac{\delta}{3} \\
\end{align}
So choose $\delta = \min\{3, \epsilon\}$ and you are done.
