University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 21


(a) The open interval is $(10/3,5)$ (b) $\delta=2/3$

Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-4|\lt\delta\Rightarrow|\frac{1}{x}-\frac{1}{4}|\lt0.05$$ 1) Find the interval around $4$ on which $|\frac{1}{x}-\frac{1}{4}|\lt0.05$ holds. Solve the inequality: $$|\frac{1}{x}-\frac{1}{4}|\lt0.05$$ $$-0.05\lt\frac{1}{x}-\frac{1}{4}\lt0.05$$ $$0.2\lt\frac{1}{x}\lt0.3$$ $$\frac{1}{0.2}\gt x\gt\frac{1}{0.3}$$ $$5\gt x\gt\frac{10}{3}$$ The open interval around $4$ is $(10/3,5)$. 2) Give a value for $\delta$ The nearer endpoint to $4$ is $10/3$, and the distance between them is $4-10/3=2/3$. So if we take $\delta=2/3$ or any smaller positive number, then $0\lt|x-4|\lt2/3$, meaning all $x$ would be placed in the interval $(10/3,5)$ so that $|\frac{1}{x}-\frac{1}{4}|\lt0.05$. In other words, $$0\lt |x-4|\lt2/3\Rightarrow|\frac{1}{x}-\frac{1}{4}|\lt0.05$$
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