How to prove this delta-epsilon proof?

Question

$$\lim_{x \rightarrow 9} \sqrt{x} = 3$$

This means we need to satisfy $|\sqrt{x} - 3| < \epsilon$ where $0 < |x - 9| < \delta$

But I don't really know how to make the two conditions match so we can link delta and epsilon together. Absolute values mess with my head and I don't know the right way to work with them.

Do I need to do something like:

$0 < |x - 9| < \delta$ turns to

$0 < |(\sqrt{x} + 3)(\sqrt{x} - 3)| < \delta$

$0 < |\sqrt{x} + 3| |\sqrt{x} - 3| < \delta$

$0 < (\sqrt{x} + 3) |\sqrt{x} - 3| < \delta$

$0 < |\sqrt{x} - 3| < \frac{\delta }{\sqrt{x} + 3} $

Answer 1

$$|x-9|<\delta\implies |\sqrt x-3||\sqrt x+3|<\delta\implies |\sqrt x-3|<\frac{\delta}{|\sqrt x+3|}=\frac{\delta}{\sqrt x+3}\le\frac{\delta}{3}=\epsilon$$ then letting $\delta=3\epsilon$ leads us to what we want.

Answer 2

By your work.

Let $\delta=3\epsilon$.

Thus, $|x-9|<3\epsilon$ gives, which you wish: $$|\sqrt{x}-3|<\frac{3\epsilon}{\sqrt{x}+3}\leq\epsilon.$$