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: 20


$(16,32)$ $\delta=7$

Work Step by Step

We are given $f(x)=\sqrt{x-7}$, $L=4$, $c=23$, and $\epsilon=1$. To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$: $$\left|\sqrt{x-7}-4 \right| \lt 1\\ -1\lt \sqrt{x-7}-4 \lt 1\\ 3\lt \sqrt{x-7} \lt 5\\ 9\lt x-7 \lt 25\\ 16 \lt x \lt 32.$$ Thus our interval is $(16,32)$. Now, to find our $\delta$, we note that $$16 \lt x \lt 32 \implies -7 \lt x-23 \lt 9.$$ Hence for $\delta=7$, $$0 \lt |x-23| \lt \delta \implies |f(x)-4| \lt \epsilon.$$
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