Question

$\underset{x\to 0}{\lim }$ $\frac{tan2x-x}{3x-sinx}$ is

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{-1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

We have, $\underset{x\to 0}{\lim }$ $\left(\frac{tan2x-x}{3x-sinx}\right)$
$=\underset{x\to 0}{\lim }$ $\left\{2x\left(\frac{tan2x}{2x}-\frac{x}{2x}\right)+x\left(\frac{3x}{x}-\frac{sinx}{x}\right)\right\}$
$=2\frac{\underset{x\to 0}{\lim }\left\{\frac{tan2x}{2x}-\frac{1}{2}\right\}}{\underset{x\to 0}{\lim }\left\{3-\frac{sinx}{x}\right\}}$
$=\frac{2\left(\underset{x\to 0}{\lim }\frac{tan2x}{2x}-\frac{1}{2}\right)}{3-\underset{x\to 0}{\lim }\left(\frac{sinx}{x}\right)}$
$=\frac{2\left(1-\frac{1}{2}\right)}{3-1}=\frac{1}{2}$