Calculus 8th Edition

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: 12

Answer

$\frac{3}{7}$

Work Step by Step

$\lim\limits_{x \to -3}\frac{x^2 + 3x}{x^2 - x - 12} = \lim\limits_{x \to -3}\frac{x (x+3)}{(x-4)(x+3)} = \lim\limits_{x \to -3}\frac{x}{(x-4)} = \frac{-3}{-3 - 4} = \frac{3}{7}$
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