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: 21

Answer

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

Work Step by Step

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