Question

$\displaystyle \lim_{x \to 0}$ $\frac{tan\,2x-x}{3x-sin\,x}$ is

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

Answer 1

Answer: (B)

We have, $\displaystyle \lim_{x \to 0}$ $\left(\frac{tan\,2x-x}{3x-sin\,x} \right)$
$=\displaystyle \lim_{x \to 0}$ $\left\{2x\left(\frac{tan\,2x}{2x}-\frac{x}{2x}\right)+x\left(\frac{3x}{x}-\frac{sin\,x}{x}\right)\right\}$
$=2 \frac{\displaystyle \lim_{x \to 0} \left\{\frac{tan\,2x}{2x}-\frac{1}{2}\right\}}{\displaystyle \lim_{x \to 0} \left\{3-\frac{sin\,x}{x}\right\}}$
$=\frac{2\left(\displaystyle \lim_{x \to 0}\frac{tan\,2x}{2x}-\frac{1}{2}\right)}{3-\displaystyle \lim_{x \to 0}\left(\frac{sin\,x}{x}\right)}$
$=\frac{2\left(1-\frac{1}{2}\right)}{3-1}=\frac{1}{2}$