Question

The value of $\underset{x\to 0}{\lim }$ $\frac{\left(1-cos2x\right)sin5x}{x^{2}sin3x}$ is

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

Answer 1

Answer: (A)

$\underset{x\to 0}{\lim }$ $\frac{\left(1-cos2x\right)sin5x}{x^{2}sin3x}$
$=\underset{x\to 0}{\lim }$ $\frac{2si{n}^{2}xsin5x}{x^{2}sin3x}$
$=\underset{x\to 0}{\lim }$$\left(\frac{2si{n}^{2}x}{x^2}\right)\frac{\left(\frac{sin5x}{5x}\right)\times 5}{\left(\frac{sin3x}{3x}\right)\times 3}$
$=\underset{x\to 0}{\lim }$ ${\left(\frac{sinx}{x}\right)}^2\times \frac{5\underset{5x\to 0}{\lim }\left(\frac{sin5x}{5x}\right)}{3\underset{3x\to 0}{\lim }\left(\frac{sin3x}{3x}\right)}$
$=\frac{2\times 5}{3}=\frac{10}{3}$