Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - Review - Exercises: 38

Answer

$-1$

Work Step by Step

$\lim _{x\rightarrow 1}\left( \dfrac {1}{x-1}+\dfrac {1}{x^{2}-3x+2}\right) =\dfrac {1}{x-1}+\dfrac {1}{\left( x-2\right) \left( x-1\right) }=\dfrac {x-2+1}{\left( x-2\right) \left( x-1\right) }=\dfrac {1}{x-2}=-1$
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