Answer
$\lim\limits_{x \to 2^-}\frac{x^2-2x}{x^2-4x+4}=-\infty$
Work Step by Step
$\lim\limits_{x \to 2^-}\frac{x^2-2x}{x^2-4x+4}=\lim\limits_{x \to 2^-}\frac{x(x-2)}{(x-2)^2}=\lim\limits_{x \to 2^-}\frac{x}{x-2}=\frac{2^-}{0^-}=-\infty$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
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 - Page 94: 40
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
