Question

$\underset{x\to \infty }{lim}\frac{\left(1-\cos 2x\right)\sin 5x}{x^2\sin 3x}$ is 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 $\frac{10}{3}$

Now,

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

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

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

$=2\times 1^2\times \frac{5}{3}$

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