Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - 2.6 Continuity - 2.6 Exercises: 31


$2\sqrt {6}$

Work Step by Step

$\lim _{x\rightarrow 4}\sqrt {\dfrac {x^{3}-2x^{2}-8x}{x-4}}=\sqrt {\dfrac {x\left( x^{2}-2x-8\right) }{x-4}}=\sqrt {\dfrac {x\left( x-4\right) \left( x+2\right) }{x-4}}=\sqrt {x\left( x+2\right) }=\sqrt {4\times \left( 4+2\right) }=2\sqrt {6}$
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