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 - Page 66: 30


$\lim\limits_{y \to 0}\frac{(5y^{3}+8y^{2})}{(3y^{4}-16y^{2})}=-0.5=-\frac{1}{2}$

Work Step by Step

Simplify: $\frac{(5y^{3}+8y^{2})}{(3y^{4}-16y^{2})} = \frac{y^{2}(5y+8)}{y^{2}(3y^{2}-16)} = \frac{(5y+8)}{(3y^{2}-16)}$ Now, $\lim\limits_{y \to 0}\frac{(5y^{3}+8y^{2})}{(3y^{4}-16y^{2})} = \lim\limits_{y \to 0}\frac{(5y+8)}{(3y^{2}-16)} = -\frac{8}{16} = -\frac{1}{2} = -0.5$
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