Answer
$\frac{1}{6}$
Work Step by Step
$\lim _{z\rightarrow 9}\dfrac {\sqrt {z}-3}{z-9}=\dfrac {\sqrt {z}-3}{\left( \sqrt {z}-3\right) \left( \sqrt {z}+3\right) }=\dfrac {1}{\sqrt {z}+3}=\dfrac {1}{6}$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 13 - Section 13.1 - Finding Limits Numerically and Graphically - 13.1 Exercises - Page 905: 14
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