Prove $\lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$ with epsilon-delta I am looking to show that $$f:(0,1)\rightarrow \mathbb{R} \lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$$ Normally for these types of problems I have been doing epsilon-delta proofs but I cannot figure out how to define my $\delta$.  
Essentially I need with finding an appropriate $\delta$. I am not looking for an application of L'Hopital's Rule. 
So far with determining a $\delta$ from $\varepsilon$ I have $$\left| \frac{\sqrt{9-x}-3}{x}+\frac{1}{6} \right|<\varepsilon $$
$$\left| \frac{-1}{\sqrt{9-x}-3}+\frac{1}{6} \right|<\varepsilon $$
$$\left| \frac{-1}{\sqrt{9-x}-3}\right|+\frac{1}{6} \leq\varepsilon $$
by the triangle inequality
$$\left| \frac{-1}{\sqrt{9-x}-3}\right| \leq\varepsilon -\frac{1}{6}$$
$$ \frac{1}{\left|\sqrt{9-x}-3\right|} \leq\varepsilon -\frac{1}{6}$$
$$ \frac{1}{\varepsilon -\frac{1}{6}} \leq\left|\sqrt{9-x}-3\right|$$
$$ \frac{1}{\varepsilon -\frac{1}{6}} \leq\left|\sqrt{9-x}\right|+3$$
$$ \left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3 \leq\left|\sqrt{9-x}\right|$$ because the square root is always positive 
$$ \left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3 \leq\sqrt{9-x}$$
$$ \left(\left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3\right)^2 \leq9-x$$
$$ 9-\left(\left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3\right)^2 \geq x$$

With 
$$\frac{\sqrt{9-x}-3}{x}=\frac{-1}{\sqrt{9-x}+3}$$
then
\begin{align}
\left|\frac{\sqrt{9-x}-3}{x}+\frac{1}{6}\right|
&= \left|\frac{-1}{\sqrt{9-x}+3}+\frac{1}{6}\right|\\
&= \left|\frac{\sqrt{9-x}-3}{6(\sqrt{9-x}+3)}\right|\\
&= \left|\frac{-x}{6(\sqrt{9-x}+3)^2}\right|\\
&\leq\dfrac{1}{54}|x|
\end{align}
