Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.7 L'Hôpital's Rule - Exercises - Page 366: 24



Work Step by Step

We have $$ \lim _{x \rightarrow 8} \frac{x^{5/3}-2x-16}{x^{1/3}-2}=\frac{32-32}{2-2}=\frac{0}{0} $$ is an intermediate form, then we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow 8} \frac{x^{5/3}-2x-16}{x^{1/3}-2}=\lim _{x \rightarrow 8} \frac{\frac{5}{3}x^{2/3}-2}{\frac{1}{3}x^{-2/3}}=\frac{\frac{20}{3}-2}{\frac{1}{12}}=80-24=56. $$
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