Answer
$\lim\limits_{x\to2^-}\dfrac{1}{\sqrt[3]{x^2-4}}=-\infty.$
Work Step by Step
$\lim\limits_{x\to2^-}\dfrac{1}{\sqrt[3]{x^2-4}}=\dfrac{1}{\sqrt[3]{(2^-)^2-4}}=\dfrac{1}{\sqrt[3]{0^-}}=-\infty.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 - Page 92: 78
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