Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.5 Exercises: 38



Work Step by Step

$\lim\limits_{x\to-\frac{1}{2}}\dfrac{6x^2+x-1}{4x^2-4x-3}=\lim\limits_{x\to-\frac{1}{2}}\dfrac{(3x-1)(2x+1)}{(2x-3)(2x+1)}$ $=\lim\limits_{x\to-\frac{1}{2}}\dfrac{3x-1}{2x-3}=\dfrac{3(-\frac{1}{2})-1}{2(-\frac{1}{2})-3}=\dfrac{5}{8}.$
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