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: 12



Work Step by Step

Since $$ \lim _{x \rightarrow -\infty} x \sin \frac{1 }{x}=-\infty\times0 $$ is an intermediate form, then we can apply L’Hôpital’s Rule as follws $$ \lim _{x \rightarrow -\infty} x\sin \frac{1 }{x}=\lim _{x \rightarrow -\infty} \frac{\sin \frac{1 }{x}}{1/x}=\lim _{x \rightarrow -\infty} \frac{-\frac{1}{x^2}\cos \frac{1 }{x}}{-\frac{1}{x^2}}=\lim _{x \rightarrow -\infty} \cos \frac{1 }{x}=1. $$
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