Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 20

Answer

$$\lim\limits_{x\to1}\Big(\frac{1}{x-1}+\frac{1}{x^2-3x+2}\Big)=-1$$

Work Step by Step

$$A=\lim\limits_{x\to1}\Big(\frac{1}{x-1}+\frac{1}{x^2-3x+2}\Big)$$ $$A=\lim\limits_{x\to1}\Big[\frac{1}{x-1}+\frac{1}{(x-1)(x-2)}\Big]$$ $$A=\lim\limits_{x\to1}\frac{(x-2)+1}{(x-1)(x-2)}$$ $$A=\lim\limits_{x\to1}\frac{x-1}{(x-1)(x-2)}$$ $$A=\lim\limits_{x\to1}\frac{1}{x-2}$$ $$A=\frac{1}{1-2}$$ $$A=-1$$
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