Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises: 40

Answer

$\lim\limits_{x \to 2^-}\frac{x^2-2x}{x^2-4x+4}=-\infty$

Work Step by Step

$\lim\limits_{x \to 2^-}\frac{x^2-2x}{x^2-4x+4}=\lim\limits_{x \to 2^-}\frac{x(x-2)}{(x-2)^2}=\lim\limits_{x \to 2^-}\frac{x}{x-2}=\frac{2^-}{0^-}=-\infty$
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