<s>[INST] 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. 
 [/INST] 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 $$|x-4| \lt 0.01$$
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