Answer
$7$
Work Step by Step
$\lim _{x\rightarrow 3}\dfrac {x^{2}+x-12}{x-3}=\dfrac {x^{2}-9+\left( x-3\right) }{x-3}=\dfrac {\left( x+3\right) \left( x-3\right) +\left( x-3\right) }{x-3}=\dfrac {\left( x-3\right) \left( x+4\right) }{x-3}=x+4=7$
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: 11
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