Chapter 3.3, Problem 44E

To determine

Find the given limit using formula $2$ .

Answer to Problem 44E

The $\underset{x\to 0}{\lim }\frac{\sin 3x\sin 5x}{x^2}=15$ .

Explanation of Solution

Given:

The given limitis $\underset{x\to 0}{\lim }\frac{\sin 3x\sin 5x}{x^2}$ .

Calculation:

Use derivative rule $\frac{d}{dx}\sin \left(kx\right)=k\cos kx$ .

  $\begin{array}{l}\underset{x\to 0}{\lim }\frac{\sin 3x\sin 5x}{x^2}\\ =\underset{x\to 0}{\lim }\frac{\sin 3x}{x}\cdot \frac{\sin 5x}{x}\\ =\underset{x\to 0}{\lim }3\cdot \frac{\sin 3x}{3x}\cdot 5\cdot \frac{\sin 5x}{5x}\\ =15\left[\underset{x\to 0}{\lim }\frac{\sin 3x}{3x}\cdot \frac{\sin 5x}{5x}\right]\\ =15\left[\left(\underset{x\to 0}{\lim }\frac{\sin 3x}{3x}\right)\cdot \left(\underset{x\to 0}{\lim }\frac{\sin 5x}{5x}\right)\right]\end{array}$

Use limit formula $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$ .

  $\begin{array}{l}=15\cdot 1\cdot 1\\ =15\end{array}$

Hence the value of the $\underset{x\to 0}{\lim }\frac{\sin 3x\sin 5x}{x^2}=15$ .