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: 19

Answer

$a-\\\lim_{x \to 4^+} \frac{x-5}{(x-4)^2}=-\infty \\ b-\\\lim_{x \to 4^-} \frac{x-5}{(x-4)^2}=-\infty \\ c-\\ \because \lim_{x \to 4^+} \frac{x-5}{(x-4)^2}=\lim_{x \to 4^-} \frac{x-5}{(x-4)^2}=-\infty \\ \therefore \lim_{x \to 4} \frac{x-5}{(x-4)^2}=-\infty \\ $

Work Step by Step

$\lim_{x \to 4^+} \frac{x-5}{(x-4)^2}=-\infty \\ (as\,x\,approach\,4\,from\,right\,\,\,the\,numerator\,(x-5)approach\,\,-1\,\\and\,(x-4)^2\,is\,positive\,and\,approach\,0 )\\ \lim_{x \to 4^-} \frac{x-5}{(x-4)^2}=-\infty \\ (as\,x\,approach\,4\,from\,left\,\,\,the\,numerator\,(x-5)approach\,\,-1\,\\and\,(x-4)^2\,is\,positive\,and\,approach\,0 )\\ \because \lim_{x \to 4^+} \frac{x-5}{(x-4)^2}=\lim_{x \to 4^-} \frac{x-5}{(x-4)^2}=-\infty \\ \therefore \lim_{x \to 4} \frac{x-5}{(x-4)^2}=-\infty \\ $
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