Using $\epsilon-\delta$ definition to show $\frac{\sqrt{9-x}-3}{x}$ as $x\to 0$

Question

Given this $\frac{\sqrt{9-x}-3}{x}-\frac{1}{6}$ I managed to reduce it to $\frac{1}{6}$ so I'm not sure if I made an error there. What I am trying to show is that there is a limit at $0$ so we have $0<|x|<\delta$ implies $\left|\frac{\sqrt{9-x}-3}{x}-\frac{1}{6}\right|<\epsilon$. I'm struggling with this a bit and I've had some trouble finding helpful references so some more thorough responses are greatly appreciated!

Answer 1

HINT:

$$\left|\frac{\sqrt{9-x}-3}{x}+\frac{1}{6}\right|=\left|\frac16-\frac{1}{\sqrt{9-x}+3}\right|=\frac{\frac16 |x|}{(\sqrt{9-x}+3)^2}$$

Answer 2

Just to make sure I'm clear here and that everything is answered thoroughly we consider

$\frac{1}{(\sqrt{9-x}+3)^2}$ we can restrict $\delta<5$ which gives

$\frac{1}{(\sqrt{9-x}+3)^2}<\frac{1}{(\sqrt{4}+3)^2}<\frac{1}{25}$

Giving

$\left| \frac{x}{150} \right|< \epsilon$

So we take $0<|x|<\min\left(5,150\epsilon\right)$