Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.4 Infinite Limits - 2.4 Exercises - Page 86: 22

Answer

$a-\\\lim\limits_{x \to -2^+ }\frac{(x-4)}{x(x+2)}=+\infty \\ b-\\\lim\limits_{x \to -2^- }\frac{(x-4)}{x(x+2)}=-\infty \\ c-\\\because \lim\limits_{x \to -2^+ }\frac{(x-4)}{x(x+2)}\neq \lim\limits_{x \to -2^- }\frac{(x-4)}{x(x+2)}\\ \therefore \lim\limits_{x \to -2 }\frac{(x-4)}{x(x+2)}\,does\,not\,exist. $

Work Step by Step

$\lim\limits_{x \to -2^+ }\frac{(x-4)}{x(x+2)}=+\infty \\ (as\,x\,approach\,-2\,from\,right\,\,\,the\,numerator\,(x-4)approach\,\,-6\,\\and\,(x)approach\,-2\,,(x+2)\,is\,positive\,and\,approach\,0 \,\\so\,the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\ \lim\limits_{x \to -2^- }\frac{(x-4)}{x(x+2)}=-\infty \\ (as\,x\,approach\,-2\,from\,left\,\,\,the\,numerator\,(x-4)approach\,\,-6\,\\and\,(x)approach\,-2\,,(x+2)\,is\,negative\,and\,approach\,0 \,\\so\,the\,denominator\,is\,postive\,and\,approach\,\,\,0)\\ \because \lim\limits_{x \to -2^+ }\frac{(x-4)}{x(x+2)}\neq \lim\limits_{x \to -2^- }\frac{(x-4)}{x(x+2)}\\ \therefore \lim\limits_{x \to -2 }\frac{(x-4)}{x(x+2)}\,does\,not\,exist. $
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