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: 19

Answer

$(3,15)$ $\delta=5$

Work Step by Step

We are given $f(x)=\sqrt{19-x}$, $L=3$, $c=10$, and $\epsilon=1$. To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$: $$\left|\sqrt{19-x}-3 \right| \lt 1\\ -1 \lt \sqrt{19-x}-3 \lt 1\\ 2\lt \sqrt{19-x} \lt 4\\ 4\lt 19-x \lt 16\\ -15 \lt -x \lt -3\\ 15\gt x \gt 3\\ 3 \lt x \lt 15.$$ Thus our interval is $(3,15)$. Now, to find our $\delta$, we note that $$3 \lt x \lt 15 \implies -7\lt x-10 \lt 5.$$ Hence for $\delta=5$, $$0 \lt |x-10| \lt \delta \implies |f(x)-3| \lt \epsilon.$$
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