Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 19

Answer

The solution is $$\lim_{x\to 0}\frac{3\sin4x}{5x}=\frac{12}{5}.$$

Work Step by Step

We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to 0}\frac{3\sin4x}{5x}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to 0}\frac{(3\sin 4x)'}{(5x)'}=\lim_{x\to 0}\frac{3\cos 4x(4x)'}{5}=\lim_{x\to 0}\frac{12\cos4x}{5}=\left[\frac{12\cdot\cos 0}{5}\right]=\frac{12}{5}.$$
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