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



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 $2.$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.