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 - Practice Exercises - Page 101: 27

Answer

$0$

Work Step by Step

The limit of the numerator $\lim_{x\to1}(3x^2+1)=4$. Thus, for the entire limit to be $\infty$, the denominator needs to be zero. Therefore, we have $\lim_{x\to1}g(x)=0$
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.