Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.3 - The Precise Definition of a Limit - Exercises 2.3 - Page 66: 39


See the proof below.

Work Step by Step

To prove $\lim\limits_{x \to 9}\sqrt {x-5}=2$, let $ c=9, f(x)=\sqrt {x-5}, L=2$. For any given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ such that for any $ x, 0\lt|x-c|\lt\delta, |f(x)-L|\lt\epsilon $. This requires $|\sqrt {x-5}-2|\lt\epsilon $, or $-\epsilon\lt \sqrt {x-5}-2\lt\epsilon $ or $2-\epsilon\lt \sqrt {x-5}\lt 2+\epsilon $, which gives $(2-\epsilon)^2-4\lt x-9\lt(2+\epsilon)^2-4$ The above process indicates that if we let $\delta $ be the smaller of $|(2-\epsilon)^2-4|$ and $|(2+\epsilon)^2-4|$, we should have for any $ x,0\lt|x-9|\lt\delta $, such that $|f(x)-2|\lt\epsilon $, which proves the limit statement.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.