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: 19

Answer

$$ \frac{3}{-5}$$

Work Step by Step

We have $$ \lim _{x \rightarrow -\infty} \frac{3x-2}{1-5x}=-\frac{\infty}{\infty}. $$ is an intermediate form, then we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow -\infty} \frac{3x-2}{1-5x}=\lim _{x \rightarrow -\infty} \frac{3}{-5}= \frac{3}{-5}. $$
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