Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 54: 16


See the proof below.

Work Step by Step

We have $$ |x^3+12-12| = |x^3-0| =x^2 |x-0|$$ Since $|x^3+12 -12| $ is a multiple of $|x-0|$, then $|x^3+12 -12| $ is arbitrarily small whenever $ x $ is sufficiently close to $0$. Hence, we get $$\lim_{x\rightarrow 0}x^3+12=12.$$
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.