Proving the limit of a radical using epsilon delta Define $f:(0,1)->R$ by $f(x)=\frac{\sqrt{9-x}-3}{x} $
I know that $f(x)=\frac{\sqrt{9-x}-3}{x}*\frac{\sqrt{9-x}+3}{\sqrt{9-x}+3}=\frac{9-x-9}{\sqrt{9-x}+3}=-\frac{1}{\sqrt{9-x}+3} $
Since $f$ is only defined in $(0,1)$, $-\frac{1}{5}<-\frac{1}{\sqrt{9-x}+3}<-\frac{1}{6} $
So can I just pick some $\delta$ close to 0?
Therefore $\lvert -\frac{1}{\sqrt{9-x}+3}+\frac{1}{6} \rvert =\lvert \frac{-6+\sqrt{9-x}+3}{6(\sqrt{9-x}+3)} \rvert =\lvert \frac{\sqrt{9-x}-3}{6\sqrt{9-x}+18}\rvert=\lvert \frac{1}{6}*\frac{\sqrt{9-x}-3}{\sqrt{9-x}+3} \rvert < \epsilon$
Take $\delta = \frac{\epsilon}{6}$
$\therefore \lvert -\frac{1}{\sqrt{9-x}+3}+\frac{1}{6} \rvert < \epsilon $ if $0<\lvert x- 0 \rvert < \delta $ for all $x \in (0,1)$

From what you wrote I do not see by what you make at the choice. Instead:
If $0 < x < 1$, then
$$
\bigg| \frac{-1}{\sqrt{9-x}+3} + \frac{1}{6} \bigg|
=
\frac{3 - \sqrt{9-x}}{6\sqrt{9-x}+18} < 3 - \sqrt{9-x} =: s_{x} \leq 3; 
$$
given any $0 < \varepsilon \leq 3$, we have $s_{x} < \varepsilon$ if $x < 9 - (3-\varepsilon)^{2}$; putting the above together, we may take $\delta := 9 - (3 - \varepsilon)^{2}$.
