Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises 2.6 - Page 97: 41

Answer

$-\infty $

Work Step by Step

When $ x $ approaches -8 from the right then $ x+8$ approaches 0 from the left because $ x+8<0$, so $$\lim _{x \rightarrow-8^{+}} \frac{2x}{ x+8}= - \infty $$
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