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: 19


(a) The interval is $(3,15)$. (b) $\delta=5$

Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-10|\lt\delta\Rightarrow|\sqrt{19-x}-3|\lt1$$ 1) Find the interval around $10$ on which $|\sqrt{19-x}-3|\lt1$ holds. Solve the inequality: $$|\sqrt{19-x}-3|\lt1$$ $$-1\lt\sqrt{19-x}-3\lt1$$ $$2\lt\sqrt{19-x}\lt4$$ Square: $$4\lt19-x\lt16$$ $$-15\lt-x\lt-3$$ $$15\gt x\gt3$$ The open interval around $10$ is $(3,15)$. 2) Give a value for $\delta$ The nearer endpoint to $10$ is $15$, and the distance between them is $15-10=5$. So if we take $\delta=5$ or any smaller positive number, then $0\lt|x-10|\lt5$, meaning all $x$ would be placed in the interval $(3,15)$ so that $|\sqrt{19-x}-3|\lt1$. In other words, $$0\lt |x-10|\lt5\Rightarrow|\sqrt{19-x}-3|\lt1$$
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