Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - 2.5 Limits at Infinity - 2.5 Exercises: 24

Answer

$-\infty$

Work Step by Step

$\lim _{x\rightarrow -\infty }\left( 2x-8+4x^{3}\right) =\dfrac {2}{x^{8}}+4x^{3}=0+4x^{3}=4\times \left( -\infty \right) =-\infty $
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