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