Question

The value of $\underset{x\to 0}{lim}\frac{2x\sin 2x+3\tan x+x^2\sin x}{\sin 3x}$ is

• A. $2$
• B. $3$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

Given : $\underset{x\to 0}{lim}\frac{2x\sin 2x+3\tan x+x^2\sin x}{\sin 3x}$
$=\underset{x\to 0}{lim}\frac{2x\times 2x\left(\frac{\sin 2x}{2x}\right)+3\times x\left(\frac{\tan x}{x}\right)+x^2\times x\left(\frac{\sin x}{x}\right)}{3x\left(\frac{\sin 3x}{3x}\right)}$
$=\underset{x\to 0}{lim}\frac{4x^2+3x+x^3}{3x}$
$=\underset{x\to 0}{lim}\frac{4x+3+x^2}{3}=\frac{3}{3}=1$