Question

The value of $\underset{x\to 0}{lim}\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}$ equal to

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

Answer 1

The correct option is A $\frac{10}{3}$

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

$=\underset{x\to 0}{lim}\frac{1-\cos 2x}{x^2}.\underset{x\to 0}{lim}\frac{\sin 5x}{\sin 3x}$

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

$=\underset{x\to 0}{lim}2{\left(\frac{\sin x}{x}\right)}^2\times \frac{5}{3}=\frac{10}{3}$