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: 21


$\left(\dfrac{10}{3}, 5\right)$ $\delta=\dfrac{2}{3}$

Work Step by Step

We are given $f(x)=\dfrac{1}{x}$, $L=\dfrac{1}{4}$, $c=4$, and $\epsilon=0.05=\dfrac{1}{20}$. To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$: $$\left|\dfrac{1}{x}-\dfrac{1}{4} \right| \lt \dfrac{1}{20}\\ -\dfrac{1}{20}\lt \dfrac{1}{x}-\dfrac{1}{4} \lt \dfrac{1}{20}\\ \dfrac{4}{20}\lt \dfrac{1}{x} \lt \dfrac{6}{20}\\ \dfrac{1}{5}\lt \dfrac{1}{x} \lt \dfrac{3}{10}\\ 5 \gt x \gt \dfrac{10}{3}\\ \dfrac{10}{3} \lt x \lt 5.$$ Thus our interval is $\left(\dfrac{10}{3}, 5\right)$. Now, to find our $\delta$, we note that $$\dfrac{10}{3} \lt x \lt 5 \implies -\dfrac{2}{3} \lt x-4 \lt 1 .$$ and $$-\dfrac{2}{3} \lt x-4 \lt \dfrac{2}{3} \implies -\dfrac{2}{3} \lt x-4 \lt 1.$$ Hence for $\delta=\dfrac{2}{3}$, $$0 \lt |x-4| \lt \delta \implies \left|f(x)-\dfrac{1}{4}\right| \lt \epsilon.$$
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