Question

Solve $\underset{x\to 0}{lim}\frac{\tan 2x-x}{3x-\sin x}$

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

Answer 1

Answer: (B)

$\frac{1}{2}$

$=\underset{x\to 0}{lim}$ $\frac{\tan 2x-x}{3x-\sin x}$

put $x=0$ , $\frac{0}{0}$

By using L'hopital rule

$=\underset{x\to 0}{lim}$ $\frac{2{\sec }^{2}2x-1}{3-\cos x}$

$=\underset{x\to 0}{lim}$ $\frac{2{\sec }^{2}0-1}{3-\cos 0}$

$=$ $\frac{2(1{)}^2-1}{3-1}$

$=$ $\frac{2-1}{2}$

$=$ $\frac{1}{2}$