Answer
$\infty$
Work Step by Step
$\lim _{x\rightarrow \infty }\dfrac {x^{2}+4x+7}{x-6}=\dfrac {\dfrac {d}{dx}\left( x^{2}+4x+7\right) }{\dfrac {d}{dx}\left( x-6\right) }=\dfrac {2x+4}{1}=2x+4=\infty $
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.7 Exercises - Page 564: 25
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