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: 46

Answer

$\lim\limits_{s\to-2}f(s)=2.$

Work Step by Step

$\lim\limits_{s\to-2^-}f(s)=\lim\limits_{s\to-2^-}(-s^2-4s-2)=-(-2^-)^2-4(-2^-)-2=2.$ $\lim\limits_{s\to-2^+}f(s)=\lim\limits_{s\to-2^+}(s^2+4s+6)=(-2^+)^2+4(-2^+)+6=2.$ Since $\lim\limits_{s\to-2^-}f(s)=\lim\limits_{s\to-2^+}f(s)\to\lim\limits_{s\to-2}f(s)$ exists and is equal to $2.$
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