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: 16

Answer

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

Work Step by Step

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