Answer
$\displaystyle \frac{1}{6}$
Work Step by Step
Use the "Strategy for Evaluating Limits Algebraically"
1. f is a closed function.
We know by Th.10.1 that it is continuous, that is, L= $\displaystyle \lim_{x\rightarrow a}f(x)$ = $f(a)$,
for all a from the domain of f.
2. Plugging, we see that $x=$3 is NOT in the domain of f.
The domain of f is $x<3$.
The form is indeterminate, 0/0 so we try simplifying.
Writing
($x-9$)=$x-3^{2}=(\sqrt{x})^{2}-3^{2}$
= ... difference of squares...
$=(\sqrt{x}+3)(\sqrt{x}-3) ...$we can reduce...
$\displaystyle \lim_{x\rightarrow 9}\frac{\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)}=\lim_{x\rightarrow 9}\frac{1}{(\sqrt{x}+3)}=$... plug...
$=\displaystyle \frac{1}{3+3}=\frac{1}{6}$
