Question

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

• A. $1$
• B. $-1$
• C. $\frac{1}{2}$
• D. none of these

Answer 1

The correct option is C $\frac{1}{2}$

Now,

$\underset{x\to 0}{lim}\frac{\tan 2x-x}{3x-\sin x}$ $\left(\frac{0}{0}\right)$ form.

Now applying L'Hospital's rule we get,

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

$=\frac{2-1}{3-1}$

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