Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises: 12

Answer

$\frac{3}{7}$

Work Step by Step

$\lim\limits_{x \to -3}\frac{x^2+3x}{x^2-x-12}=\frac{(-3)^2+3(-3)}{(-3)^2-(-3)-12}=\frac{0}{0}$ First step is to plug in the x value but since it comes out to 0/0 we have to do algebra. $\lim\limits_{x \to -3}\frac{(x)(x+3)}{(x-4)(x+3)}=\lim\limits_{x \to -3}\frac{x}{x-4}=\frac{-3}{-3-4}=\frac{3}{7}$
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