Question

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

• A. $10/3$
• B. $3/10$
• C. $6/5$
• D. $5/6$

Answer 1

Answer: (A)

$10/3$

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

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

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

$=\frac{10}{3}$.