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 
$$ \left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3 \leq\sqrt{9-x}$$
$$ \left(\left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3\right)^2 \leq9-x$$
$$ 9-\left(\left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3\right)^2 \geq x$$
 A: 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}
A: Consider that $$\frac{\sqrt{9-x}-3}{x}+\frac{1}{6} = \frac{x+6\sqrt{9-x}-18}{6x} = -\frac{\left(\sqrt{9-x}-3\right)^2}{6x}$$ If $\lim_{x\to 0^+} \frac{f(x)}{x} = c$, then for $\epsilon > 0$, there is a $\delta > 0$ such that $$\left\lvert \frac{f(x)}{x}-c\right\rvert = \left\lvert \frac{f(x)-cx}{x}\right\rvert < \epsilon \Leftrightarrow \lvert f(x)-cx\rvert < \epsilon x$$ for $0 < x < \delta$. Therefore, for $0 < x < \delta$, \begin{align*} \left\lvert \frac{f(x)^2}{x}\right\rvert &= \left\lvert \frac{((f(x)-cx)+cx)^2}{x}\right\rvert \\ &= \left\lvert \frac{(f(x)-cx)^2+2(f(x)-cx)cx+c^2x^2}{x}\right\rvert \\ &\leq \frac{\lvert f(x)-cx\rvert^2}{\lvert x\rvert}+\frac{2c\lvert f(x)-cx\rvert x}{x}+c^2x \\ &< (c+\epsilon)^2x \end{align*} so $\lim_{x\to 0^+} \frac{f(x)^2}{x} = 0$. We get the same result for $\lim_{x\to 0^-} \frac{f(x)}{x} = d$, so if the directional limits of $\frac{f(x)}{x}$ exist as $x\to 0$, then $\lim_{x\to 0} \frac{f(x)^2}{x} = 0$. Now, if you can reason that the directional limits of $\frac{\sqrt{9-x}-3}{x}$ exist as $x\to 0$,$^1$ then you will have $$\lim_{x\to 0} \frac{\sqrt{9-x}-3}{x} = -\frac{1}{6}$$

$^1$This follows from the facts that $f(x) = \frac{\sqrt{9-x}-3}{x}$ is monotonic on $(-\infty, 0)$ and $(0, 9)$, as $f'(x) = -\frac{\left(\sqrt{9-x}-3\right)^2}{2x^2\sqrt{9-x}} < 0$ for $x\neq 0, 9$, and that $f(x)$ is bounded, i.e. $-\frac{1}{3}\leq f(x) < 0$. The directional limits of monotonic, bounded functions always exist.
