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: 8

Answer

$$-6$$

Work Step by Step

Since we have $$ \lim _{x \rightarrow 0} \frac{x^3}{\sin x -x}=\frac{0}{0} $$ then we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow 0} \frac{3x^2}{\cos x -1}=\frac{0}{0}. $$ We have to apply L’Hôpital’s Rule again $$ \lim _{x \rightarrow 0} \frac{6x}{-\sin x }=\frac{0}{0}. $$ Since the form is intermediate, we have to apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow 0} \frac{6}{-\cos x }=-6. $$
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