help solving for delta in epsilon delta proof $$\lim_{x\to 9} \frac{1}{\sqrt{x}} = \frac{1}{3} $$ 
so the two statements are
$$0<|x-9|<\delta$$
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|<\epsilon$$
I've tried to multiply by the conjegate to get
$$\frac{\frac{1}{x}-\frac{1}{9}}{\sqrt{\frac{1}{x}}+\frac{1}{3}}<\epsilon$$
I'm unsure where to go from here to get that x-9 factor i need
 A: Note that$$\left\lvert\frac1{\sqrt x}-\frac13\right\rvert=\frac{\left\lvert3-\sqrt x\right\rvert}{3\sqrt x}=\frac{\lvert9-x\rvert}{3\sqrt x\left(3+\sqrt x\right)}.$$Now, if $\lvert x-9\rvert<5$, then $x>4$ and therefore $\sqrt x>2$. Also, $3+\sqrt x>5$. So $3\sqrt x\left(3+\sqrt x\right)>30$. Therefore, if you take $\delta=\min\left\{5,30\varepsilon\right\}$, then$$\lvert x-9\rvert<\delta\implies\left\lvert\frac1{\sqrt x}-\frac13\right\rvert=\frac{\lvert9-x\rvert}{3\sqrt x\left(3+\sqrt x\right)}<\frac{\lvert9-x\rvert}{30}<\varepsilon.$$
A: We have that
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|=\left|\frac{\sqrt{x}-3}{3\sqrt{x}}\right|=\left|\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{3\sqrt{x}(\sqrt{x}+3)}\right|=\left|\frac{x-9}{3\sqrt{x}(\sqrt{x}+3)}\right|$$
then assuming wlog $|x-9|<1 \implies 3\sqrt{x}(\sqrt{x}+3)\ge 3\sqrt{8}(\sqrt{8}+3)$ therefore 
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|=\left|\frac{x-9}{3\sqrt{x}(\sqrt{x}+3)}\right|\le\frac{|x-9|}{ 3\sqrt{8}(\sqrt{8}+3)}$$
and it suffices to assume
$$\delta<3\sqrt{8}(\sqrt{8}+3)\epsilon $$
to have
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|<\epsilon$$
A: $\epsilon >0$ be given.
$ |\dfrac{1}{√x}-1/3|=|\dfrac{3-√x}{3√x}|=$
$|\dfrac{9-x}{(3+√x)3√x}|\lt |\dfrac{9-x}{x}|$;
Consider $|x-9|<1$, then $8<x<10$;
Choose $\delta = \min(1, 8\epsilon)$;
Then $|x-9|<\delta$ implies 
$|\dfrac{1}{√x}-1/3| <$
$|\dfrac{9-x}{x}| <(1/8)\delta \lt \epsilon.$
