Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 140: 76

Answer

(a) $x \gt 10^8$ (b) $\lim\limits_{x \to \infty}\frac{1}{\sqrt{x}} = 0$

Work Step by Step

(a) $\frac{1}{\sqrt{x}} \lt 0.0001$ $\sqrt{x} \gt \frac{1}{0.0001}$ $\sqrt{x} \gt 10^4$ $x \gt 10^8$ (b) Let $f(x) =\frac{1}{\sqrt{x}}$ This function is defined on the interval $(0, \infty)$ Let $\epsilon \gt 0$ be given. Let $N = (\frac{1}{\epsilon})^2$ Suppose that $x \gt N$ Then: $\vert \frac{1}{\sqrt{x}} - 0\vert \lt \vert \frac{1}{\sqrt{N}}\vert = \frac{1}{\sqrt{(\frac{1}{\epsilon})^2}} = \epsilon$ Therefore, $\lim\limits_{x \to \infty}\frac{1}{\sqrt{x}} = 0$
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