## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 2 - Limits - 2.6 Trigonometric Limits - Exercises - Page 77: 50

#### Answer

$$-4$$

#### Work Step by Step

Since $$\sin 3\theta =3\sin \theta -4\sin^3\theta,$$ then we have \begin{align*} \lim _{\theta \rightarrow 0} \frac{ \sin 3\theta -3\sin \theta }{ \theta^3}&= \lim _{\theta \rightarrow 0} \frac{ -4\sin^3 \theta }{ \theta^3}\\ &=-4 \lim _{\theta \rightarrow 0} \frac{ \sin^3 \theta }{ \theta^3}\\ &= -4 \left(\lim _{\theta \rightarrow 0} \frac{ \sin \theta }{ \theta}\right)^3\\ &= -4. \end{align*} Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1$.

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