Answer
$\dfrac {1}{6}$
Work Step by Step
$\lim _{x\rightarrow 9}\dfrac {\sqrt {x}-3}{x-9}=\dfrac {\sqrt {x}-3}{\sqrt {\left( x\right) }^{2}-3^{2}}=\dfrac {\sqrt {x}-3}{\left( \sqrt {x}-3\right) \left( \sqrt {x}+3\right) }=\dfrac {1}{\sqrt {x}+3}=\dfrac {1}{\sqrt {9}+3}=\dfrac {1}{6}$
