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 - 1.5 Exercises: 41

Answer

$\lim\limits_{x\to-4^-}(x^2+\dfrac{2}{x+4})=-\infty.$

Work Step by Step

$\lim\limits_{x\to-4^-}(x^2+\dfrac{2}{x+4})\to$ $\lim\limits_{x\to-4^-}x^2=(-4^-)^2=16.$ $\lim\limits_{x\to-4^-}\dfrac{2}{x+4}=\dfrac{2}{-4^-+4}=\dfrac{2}{0^-}=-\infty.$ By Theorem $1.15:$ $\lim\limits_{x\to-4^-}(x^2+\dfrac{2}{x+4})=-\infty.$
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