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: 42



Work Step by Step

$$A=\lim_{x\to-5^-}\frac{3x}{2x+10}=\lim_{x\to-5^-}\frac{(3x+15)-15}{2x+10}=\lim_{x\to-5^-}\frac{3(x+5)-15}{2(x+5)}$$ $$A=\lim_{x\to-5^-}\Big(\frac{3}{2}-\frac{15}{2(x+5)}\Big)=\frac{3}{2}-\lim_{x\to-5^-}\frac{15}{2x+10}$$ As $x\to-5^-$, $2x+10$ approaches $0$ from the left, where $(2x+10)\lt 0$, so $15/(2x+10)$ will approach $-\infty$. Therefore, $$A=\frac{3}{2}-(-\infty)=\infty$$
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