Answer
$\frac{3}{7}$
Work Step by Step
$\lim\limits_{x \to -3}\frac{x^2 + 3x}{x^2 - x - 12} = \lim\limits_{x \to -3}\frac{x (x+3)}{(x-4)(x+3)} = \lim\limits_{x \to -3}\frac{x}{(x-4)} = \frac{-3}{-3 - 4} = \frac{3}{7}$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 1 - Functions and Limits - 1.5 The Limit of a Function - 1.5 Exercises - Page 60: 12
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
