Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises: 44

Answer

$$\lim\limits_{x\to0}\frac{\sin3x\sin5x}{x^2}=15$$

Work Step by Step

$$A=\lim\limits_{x\to0}\frac{\sin3x\sin5x}{x^2}$$ Again, we would try to change the formula into $\frac{\sin3x}{3x}$ and $\frac{\sin5x}{5x}$ $$A=\lim\limits_{x\to0}\frac{\sin3x}{x}\times\lim\limits_{x\to0}\frac{\sin5x}{x}$$ $$A=3\lim\limits_{x\to0}\frac{\sin3x}{3x}\times5\lim\limits_{x\to0}\frac{\sin5x}{5x}$$ $$A=3\times1\times5\times1$$ $$A=15$$
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