University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 67: 56

Answer

a) 1 b) 0 c) $\frac{16}{3}$

Work Step by Step

a) $\lim\limits_{x \to -2}(p(x)+r(x)+s(x)) = 4+0-3 = 1$ b) $\lim\limits_{x \to -2}p(x).r(x).s(x) = 4(0)(-3) = 0$ c) $\lim\limits_{x \to -2}\frac{-4p(x)+5r(x)}{s(x)} = \frac{-4(4)+5(0)}{-3}=\frac{16}{3}$
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.