Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises: 64

Answer

$\dfrac {1}{2}$

Work Step by Step

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