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 66: 24

Answer

$\lim\limits_{x \to -3} \frac{(x+3)}{x^{2}+4x+3} == -\frac{1}{2} = -0.5$

Work Step by Step

Simplify: $\frac{(x+3)}{x^{2}+4x+3} = \frac{(x+3)}{(x+3)(x+1)} = \frac{1}{(x+1)}$ $\lim\limits_{x \to -3} \frac{(x+3)}{x^{2}+4x+3} =\lim\limits_{x \to -3} \frac{1}{x+1}$ $ = -\frac{1}{2} = -0.5$
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.