Question

$\underset{x\to 0}{\lim }$ $\frac{\left(1-\cos 2x\right)\sin 5x}{x^2\sin 3x}$ equals

• A. $\frac{10}{3}$
• B. $\frac{3}{10}$
• C. $\frac{6}{5}$
• D. $\frac{5}{6}$

Answer 1

Answer: (A)

We have, $\underset{x\to 0}{\lim }\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}$
$=\underset{x\to 0}{\lim }\frac{1-\cos 2x}{x^2}\underset{x\to 0}{\lim }\frac{\sin 5x}{\sin 3x}$
$=\underset{x\to 0}{\lim }\cdot \frac{2{\sin }^{2}x}{x^2}\underset{x\to 0}{\lim }\frac{{\sin }^{2}5x}{5x}\frac{3x}{\sin 3x}\cdot \frac{5x}{3x}$
$=\underset{x\to 0}{\lim }2\left(\frac{\sin x}{x}\right)\cdot \frac{5}{3}$
$=\frac{2\times 5}{3}=\frac{10}{3}$