Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 140: 77

Answer

$\lim\limits_{x \to -\infty}\frac{1}{x} = 0$

Work Step by Step

Let $f(x) =\frac{1}{x}$ This function is defined on the interval $(-\infty, 0)$ Let $\epsilon \gt 0$ be given. Let $N = -\frac{1}{\epsilon}$ Suppose that $x \lt N$ Then: $\vert \frac{1}{x} - 0\vert \lt \vert \frac{1}{N}\vert = \vert \frac{1}{(-\frac{1}{\epsilon})} \vert = \epsilon$ Therefore, $\lim\limits_{x \to -\infty}\frac{1}{x} = 0$
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