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: 22


$(\sqrt{2.9}, \sqrt{3.1})$ $\delta=\sqrt{3.1}-\sqrt{3}$

Work Step by Step

We are given $f(x)=x^2$, $L=3$, $c=\sqrt{3}$, and $\epsilon=0.1$. To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$: $$\left|x^2-3 \right| \lt 0.1\\ -0.1 \lt x^2-3 \lt 0.1\\ 2.9\lt x^2 \lt 3.1\\ \sqrt{2.9}\lt x \lt \sqrt{3.1} .$$ Thus our interval is $(\sqrt{2.9}, \sqrt{3.1})$. Now, to find our $\delta$, we note that $$\sqrt{2.9} \lt x \lt \sqrt{3.1} \implies \sqrt{2.9}-\sqrt{3} \lt x-\sqrt{3} \lt \sqrt{3.1}-\sqrt{3}.$$ Notice $\sqrt{2.9-\sqrt{3}} \approx -0.0291$ and $\sqrt{3.1}-\sqrt{3} \approx 0.0286$. This means $|\sqrt{3.1}-\sqrt{3}| \lt |\sqrt{2.9}-\sqrt{3}|$. Thus, $$-(\sqrt{3.1}-\sqrt{3}) \lt x-\sqrt{3} \lt \sqrt{3.1}-\sqrt{3}\\ \implies \sqrt{2.9}-\sqrt{3} \lt x-\sqrt{3} \lt \sqrt{3.1}-\sqrt{3}.$$ Hence for $\delta=\sqrt{3.1}-\sqrt{3}$, $$0 \lt |x-\sqrt{3}| \lt \delta \implies |f(x)-3| \lt \epsilon.$$
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.