## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 2 - Limits - 2.6 Trigonometric Limits - Exercises - Page 77: 42

#### Answer

$$\frac{2}{3}$$

#### Work Step by Step

\begin{align*} \lim _{x\rightarrow 0} \frac{ \sin 5x\sin 2x}{\sin 3x\sin 5x}&=\lim _{x\rightarrow 0} \frac{ \sin 2x}{\sin 3x}\\ &= \lim _{x\rightarrow 0} \frac{2}{3} \frac{ \sin 2x}{2x} \frac{ 3x}{\sin 3x} \\ &= \frac{2}{3} \lim _{2x\rightarrow 0}\frac{ \sin 2x}{2x} \lim _{3x\rightarrow 0}\frac{ 3x}{\sin 3x}\\ &= \frac{2}{3}. \end{align*} Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1.$

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