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: 29



Work Step by Step

We have $$ \lim _{x \rightarrow 3^+} \frac{x-4}{x^2-9}=\lim _{x \rightarrow 3^+} \frac{x-4}{(x+3)(x-3)}=- \infty. $$ We know that the limit from the right will be negative infinity because the numerator will be negative (since 3 is less than 4), while the denominator will be positive. The denominator will be arbitrarily close to zero, producing an infinite result.
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.