University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises: 16

Answer

$\lim\limits_{s \to \frac{2}{3}} (8-3s )(2s-1) = 2$

Work Step by Step

$\lim\limits_{s \to \frac{2}{3}} (8-3s )(2s-1) = \lim\limits_{s \to 0.6\bar6} (-6s^{2} +19s -8)$ = -$\lim\limits_{s \to 0.6\bar6}6s^{2} + 19\lim\limits_{s \to 0.6\bar6}s -\lim\limits_{s \to 0.6\bar6} 8$ = $-6(0.6\bar6)^{2} + 19*0.6\bar6 - 8$ = 2
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