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: 17

Answer

$\lim\limits_{x \to -\frac{1}{2}}4x(3x + 4)^{2} =-12.5=-\frac{25}{2}$

Work Step by Step

$\lim\limits_{x \to -\frac{1}{2}}4x(3x + 4)^{2} = \lim\limits_{x \to -0.5}(36x^{3} + 96x^{2} + 64x)$ = $36\lim\limits_{x \to -0.5}x^{3} + 96\lim\limits_{x \to -0.5}x^{2} + 64\lim\limits_{x \to -0.5}x$ = $36(-0.5)^{3} + 96(-0.5)^{2} + 64(-0.5)$ = -12.5
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