Answer
$\lim\limits_{x\to2^-}\dfrac{x^2}{x^2+4}=\dfrac{1}{2}.$
Work Step by Step
$\lim\limits_{x\to2^-}\dfrac{x^2}{x^2+4}=\dfrac{(2^-)^2}{(2^-)^2+4}=\dfrac{4}{4+4}=\dfrac{1}{2}.$
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 - 1.5 Exercises - Page 88: 36
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