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: 44



Work Step by Step

$$A=\lim_{x\to0}\frac{-1}{x^2(x+1)}=\frac{-1}{\lim_{x\to0}(x^2)(0+1)}=\frac{-1}{\lim_{x\to0}(x^2)}$$ As $x\to0$, $x$ approaches $0$ and $x^2$ approaches $0$ as well. Since $x^2\ge0$ for all $x$, $-1/x^2\lt0$, so $-1/x^2$ will approach $-\infty$. Therefore, $$A=-\infty$$
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