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 - Review Exercises: 76

Answer

$\lim\limits_{x\to-1^-}\dfrac{x+1}{x^4-1}=-\dfrac{1}{4}$.

Work Step by Step

$\lim\limits_{x\to-1^-}\dfrac{x+1}{x^4-1}=\lim\limits_{x\to-1^-}\dfrac{x+1}{(x^2-1)(x^2+1)}$ $=\lim\limits_{x\to-1^-}\dfrac{(x+1)}{(x+1)(x-1)(x^2+1)}=\lim\limits_{x\to-1^-}\dfrac{1}{(x-1)(x^2+1)}$ $=\dfrac{1}{((-1^-)-1)((-1^-)^2+1)}=-\dfrac{1}{4}.$
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