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 - Page 76: 27



Work Step by Step

$\lim _{x\rightarrow 1}\dfrac {5x^{2}+6x+1}{8\times -4}=\dfrac {5\times 1^{2}+6\times 1+1}{8\times 1-4}=\dfrac {12}{8-4}=\dfrac {12}{4}=3$
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