Answer
$-\frac{1}{2}$
Work Step by Step
$\lim _{x\rightarrow 0}\left( \dfrac {1}{x}+\dfrac {2}{x^{2}-2x}\right) =\lim _{x\rightarrow 0}\dfrac {2+x-2}{x^{2}-2x}=\dfrac {x}{x^{2}-2x}=\dfrac {1}{x-2}=-\dfrac {1}{2}$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 13 - Section 13.1 - Finding Limits Numerically and Graphically - 13.1 Exercises - Page 905: 16
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