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: 16


$(-2.01,-1.99)$ $\delta=0.01$

Work Step by Step

We are given $f(x)=2x-2$, $L=-6$, $c=-2$, and $\epsilon=0.02$. To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$: $$\left|2x-2-(-6) \right| \lt 0.02\\ \left|2x+4\right| \lt 0.02 -0.02 \lt 2x+4 \lt 0.02\\ -0.01\lt x+2 \lt 0.01\\ -2.01 \lt x \lt -1.99.$$ Thus our interval is $(-2.01,-1.99)$. Now, to find our $\delta$, we note that $$-2.01 \lt x \lt -1.99 \implies -0.01 \lt x-(-2) \lt 0.01.$$ Hence for $\delta=0.01$, $$0 \lt |x-(-2)| \lt \delta \implies |f(x)-(-6)| \lt \epsilon.$$
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.