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: 34

Answer

$\lim\limits_{v \to 2} \frac{v^{3}-8}{v^{4}-16}= \frac{3}{8} = 0.375$

Work Step by Step

Simplify: $\frac{v^{3}-8}{v^{4}-16} = \frac{(v-2)(v^{2}+2v+4)}{(v+2)(v-2)(v^{2}+4)} = \frac{(v^{2}+2v+4)}{(v+2)(v^{2}+4)}$ Now, $\lim\limits_{v \to 2} \frac{v^{3}-8}{v^{4}-16}= \lim\limits_{v \to 2} \frac{(v^{2}+2v+4)}{(v+2)(v^{2}+4)} = \frac{(2^{2} + 2(2)+ 4)}{(2+2)(2^{2}+4)} = \frac{12}{32} = \frac{3}{8} = 0.375$
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