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: 18


(a) The interval is $(0.16,0.36)$ (b) $\delta=0.09$

Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-\frac{1}{4}|\lt\delta\Rightarrow|\sqrt{x}-\frac{1}{2}|\lt0.1$$ 1) Find the interval around $0$ on which $|\sqrt{x}-\frac{1}{2}|\lt0.1$ holds. Solve the inequality: $$|\sqrt{x}-\frac{1}{2}|\lt0.1$$ $$-0.1\lt\sqrt x-\frac{1}{2}\lt0.1$$ $$0.4\lt\sqrt x\lt0.6$$ Square: $$0.16\lt x\lt0.36$$ The open interval around $1/4$ is $(0.16,0.36)$. 2) Give a value for $\delta$ The nearer endpoint to $0$ is $0.16$, and the distance between them is $1/4-0.16=0.09$. So if we take $\delta=0.09$ or any smaller positive number, then $0\lt|x-1/4|\lt0.09$, meaning all $x$ would be placed in the interval $(0.16,0.36)$ so that $|\sqrt{x}-\frac{1}{2}|\lt0.1$. In other words, $$0\lt |x-\frac{1}{4}|\lt0.09\Rightarrow|\sqrt{x}-\frac{1}{2}|\lt0.1$$
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