Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises - Page 102: 12


$\lim\limits_{x \to -3}\dfrac{x^{2}+3x}{x^{2}-x-12}=\dfrac{3}{7}$

Work Step by Step

First apply direct substitution: $\lim\limits_{x \to -3}\dfrac{x^{2}+3x}{x^{2}-x-12}=\dfrac{(-3)^{2}+3(-3)}{(-3)^{2}-(-3)-12}=\dfrac{0}{0}$ $(Indeterminate$ $Form)$ Apply factorization to the numerator and to the denominator: $\lim\limits_{x \to -3}\dfrac{x^{2}+3x}{x^{2}-x-12}=\lim\limits_{x \to -3}\dfrac{x(x+3)}{(x-4)(x+3)}=\lim\limits_{x \to -3}\dfrac{x}{x-4}=...$ $...=\dfrac{-3}{-3-4}=\dfrac{-3}{-7}=\dfrac{3}{7}$
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