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 - 1.4 Exercises - Page 80: 63


The function is continuous when $a=2.$

Work Step by Step

$\lim\limits_{x\to2^-}f(x)=(2^-)^3=8.$ $\lim\limits_{x\to2^+}f(x)=a(2^+)^2=4a.$ For the function to be continuous, $\lim\limits_{x\to2^+}f(x)=\lim\limits_{x\to2^-}f(x)\to$ $4a=8\to a=2.$
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