University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 108: 40



Work Step by Step

$$A=\lim_{x\to3^+}\frac{1}{x-3}$$ As $x\to3^+$, $x-3$ approaches $0$ from the right, where $(x-3)\gt 0$, so $1/(x-3)$ will approach $\infty$. Therefore, $$A=\lim_{x\to3^+}\frac{1}{x-3}=\infty$$
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.