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: 43

Answer

$\lim\limits_{x\to2}f(x)=0.$

Work Step by Step

$\lim\limits_{x\to2^-}f(x)=\lim\limits_{x\to2^-}(x-2)^2=(2^--2)^2=0.$ $\lim\limits_{x\to2^+}f(x)=\lim\limits_{x\to2^+}(2-x)=(2-2^+)=0.$ Since $\lim\limits_{x\to2^+}f(x)=\lim\limits_{x\to2^-}f(x)\to\lim\limits_{x\to2}f(x)$ exists and is equal to $0.$
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