University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 108: 41



Work Step by Step

$$A=\lim_{x\to-8^+}\frac{2x}{x+8}=\lim_{x\to-8^+}\frac{(2x+16)-16}{x+8}$$ $$A=\lim_{x\to-8^+}\Big(2-\frac{16}{x+8}\Big)=2-\lim_{x\to-8^+}\frac{16}{x+8}$$ As $x\to-8^+$, $x+8$ approaches $0$ from the right, where $(x+8)\gt 0$, so $16/(x+8)$ will approach $\infty$. Therefore, $$A=2-\infty=-\infty$$
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