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: 20

Answer

$a-\\\lim\limits_{x \to 1^+ }\frac{x-2}{(x-1)^3}=-\infty \\ b-\\\lim\limits_{x \to 1^- }\frac{x-2}{(x-1)^3}=+\infty \\ c-\\\because \lim\limits_{x \to 1^+ }\frac{x-2}{(x-1)^3}\neq \lim\limits_{x \to 1^- }\frac{x-2}{(x-1)^3}\\ \therefore \lim\limits_{x \to 1 }\frac{x-2}{(x-1)^3} \,does\,not\,exist\\ $

Work Step by Step

$\lim\limits_{x \to 1^+ }\frac{x-2}{(x-1)^3}=-\infty \\ (as\,x\,approach\,1\,from\,right\,\,\,the\,numerator\,(x-2)approach\,\,-1\,\\and\,(x-1)^3\,is\,positive\,and\,approach\,0 )\\ \lim\limits_{x \to 1^- }\frac{x-2}{(x-1)^3}=+\infty \\ (as\,x\,approach\,1\,from\,left\,\,\,the\,numerator\,(x-2)approach\,\,-1\,\\and\,(x-1)^3\,is\,negative\,and\,approach\,0 )\\ \because \lim\limits_{x \to 1^+ }\frac{x-2}{(x-1)^3}\neq \lim\limits_{x \to 1^- }\frac{x-2}{(x-1)^3}\\ \therefore \lim\limits_{x \to 1 }\frac{x-2}{(x-1)^3} \,does\,not\,exist\\ $
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