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: 40

Answer

$4$

Work Step by Step

$\lim _{x\rightarrow 3}\dfrac {x^{2}-2x-3}{x-3}=\dfrac {x^{2}-2x-9+6}{x-3}=\dfrac {x^{2}-9-\left( 2x-6\right) }{x-3}=\dfrac {\left( x-3\right) \left( x+3\right) -2\left( x-3\right) }{x-3}=\dfrac {\left( x-3\right) \left( x+1\right) }{x-3}=x+1=3+1=4$
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