Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises - Page 307: 13


The solution is $$\lim_{x\to 2}\frac{x^2-2x}{8-6x+x^2}=-1$$

Work Step by Step

We will evaluate this limit using L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule" $$\lim_{x\to 2}\frac{x^2-2x}{8-6x+x^2}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to 2}\frac{(x^2-2x)'}{(8-6x+x^2)'}=\lim_{x\to 2}\frac{2x-2}{-6+2x}=\left[\frac{2\cdot 2-2}{-6+2\cdot 2}\right] =-1.$$
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