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: 24

Answer

$a-\\\lim\limits_{x \to -2^+}\frac{x^3-5x^2+6x}{x^4-4x^2}=+\infty\\ b-\\ \lim\limits_{x \to -2^-}\frac{x^3-5x^2+6x}{x^4-4x^2}=-\infty \\ c-\\ \because \lim\limits_{x \to -2^+}\frac{x^3-5x^2+6x}{x^4-4x^2}\neq \lim\limits_{x \to -2^-}\frac{x^3-5x^2+6x}{x^4-4x^2}\\ \therefore \lim\limits_{x \to -2}\frac{x^3-5x^2+6x}{x^4-4x^2}\,does\,not\,exist. $ $ d-\\\lim\limits_{x \to 2}\frac{x^3-5x^2+6x}{x^4-4x^2}=\frac{-1}{8}$

Work Step by Step

$a-\\\lim\limits_{x \to -2^+}\frac{x^3-5x^2+6x}{x^4-4x^2}=\lim\limits_{x \to -2^+}\frac{x(x-3)(x-2)}{x^2(x-2)(x+2)} =\lim\limits_{x \to -2^+}\frac{x-3}{x(x+2)}\\ as\,x\,approach\,-2\,from\,right\,\,\,the\,numerator\,(x-3)\,approach\,\,-5\,\\and\,,x\,approach\,-2\,\,,(x+2),is\,positive\,and\,approach\,0 \,\\so\,the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\ so\,\lim\limits_{x \to -2^+}\frac{x-3}{x(x+2)}=+\infty \\ b-\\\lim\limits_{x \to -2^-}\frac{x^3-5x^2+6x}{x^4-4x^2}=\lim\limits_{x \to -2^-}\frac{x(x-3)(x-2)}{x^2(x-2)(x+2)}=\lim\limits_{x \to -2^-}\frac{x-3}{x(x+2)}\\ as\,x\,approach\,-2\,from\,left\,\,\,the\,numerator\,(x-3)\,approach\,\,-5\,\\and\,,x\,approach\,-2\,\,,(x+2),is\,negative\,and\,approach\,0 \,\\so\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\ so\,\lim\limits_{x \to -2^-}\frac{x-3}{x(x+2)}=-\infty \\ c-\\\because \lim\limits_{x \to -2^+}\frac{x^3-5x^2+6x}{x^4-4x^2}\neq \lim\limits_{x \to -2^-}\frac{x^3-5x^2+6x}{x^4-4x^2}\\ \therefore \lim\limits_{x \to -2}\frac{x^3-5x^2+6x}{x^4-4x^2}\,does\,not\,exist. $ $d-\\\lim\limits_{x \to 2}\frac{x^3-5x^2+6x}{x^4-4x^2}=\lim\limits_{x \to 2}\frac{x(x-3)(x-2)}{x^2(x-2)(x+2)}\\ =\lim\limits_{x \to 2}\frac{x-3}{x(x+2)}=\frac{2-3}{2(2+2)}=\frac{-1}{8} $
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