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 67: 33


$\lim\limits_{u \to 1}\frac{u^{4}-1}{u^{3}-1} = \frac{4}{3} = 1.33$

Work Step by Step

Simplify: $\frac{u^{4} - 1^{4}}{u^{3}-1} =\frac{ (u+1)(u-1)(u^{2}+1^{2})}{(u-1)(u^{2}+u+1)} = \frac{ (u+1)(u^{2}+1^{2})}{(u^{2}+u+1)}$ Now, $\lim\limits_{u \to 1}\frac{u^{4}-1}{u^{3}-1} = \lim\limits_{u \to 1}\frac{ (u+1)(u^{2}+1^{2})}{(u^{2}+u+1)} = \frac{(1+1)(1+1)}{1+1+1} = \frac{4}{3} = 1.33$
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