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: 75

Answer

(a) $x \gt 100$ (b) $\lim\limits_{x \to \infty}\frac{1}{x^2} = 0$

Work Step by Step

(a) $\frac{1}{x^2} \lt 0.0001$ $x^2 \gt \frac{1}{0.0001}$ $x^2 \gt 10,000$ $x \gt 100$ (b) Let $f(x) =\frac{1}{x^2}$ This function is defined on the interval $(0, \infty)$ Let $\epsilon \gt 0$ be given. Let $N = \sqrt{\frac{1}{\epsilon}}$ Suppose that $x \gt N$ Then: $\vert \frac{1}{x^2} - 0\vert \lt \vert \frac{1}{N^2}\vert = \frac{1}{(\sqrt{\frac{1}{\epsilon}})^2} = \epsilon$ Therefore, $\lim\limits_{x \to \infty}\frac{1}{x^2} = 0$
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