Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises: 45

Answer

$-12$

Work Step by Step

$\lim _{x\rightarrow -1}\dfrac {\left( 2x-1\right) ^{2}-9}{x+1}=\dfrac {\left( 2x-1\right) ^{2}-3^{2}}{x+1}=\dfrac {\left( 2x-1+3\right) \left( 2x-1-3\right) }{x+1}=\dfrac {\left( 2x+2\right) \left( 2x-4\right) }{x+1}=\dfrac {2\left( x+1\right) \left( 2x-4\right) }{x+1}=2(2x-4)=4x-8=4\times (-1)-8=-12$
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