University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 75: 10

Answer

$\delta=0.39$

Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-3|\lt\delta\Rightarrow|f(x)-4|\lt0.2$$ Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $3.8$ and $4.2$ (which means $|f(x)-4|\lt0.2$), $x$ must be placed between $2.61$ and $3.41$. - Calculate the distance from $3$ to $2.61$ and $3.41$: $3-2.61=0.39$ and $3.41-3=0.41$. So we would go with the nearer endpoint, which is $2.61$ and take $\delta$ to be $0.39$ ( or any smaller positive number is fine). Therefore, $$0\lt |x-3|\lt0.39\Rightarrow|f(x)-4|\lt0.2$$
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.