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: 23

Answer

The solution is $$\lim_{x\to0}\frac{1-\cos3x}{8x^2}=\frac{9}{16}.$$

Work Step by Step

We will use L'Hopital's rule to calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to0}\frac{1-\cos3x}{8x^2}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(1-\cos3x)'}{(8x^2)'}=\lim_{x\to0}\frac{-(-\sin3x)(3x)'}{8\cdot2x}=\lim_{x\to0}\frac{3\sin3x}{16x}=\left[\frac{3\cdot\sin(3\cdot0)}{16\cdot0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(3\sin3x)'}{(16x)'}=\lim_{x\to0}\frac{3\sin3x(3x)'}{16}=\lim_{x\to0}\frac{9\cos3x}{16}=\frac{9\cos(3\cdot0)}{16}=\frac{9}{16}.$$
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