Question

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

• A. 2
• B.
• C. -1
• D.

Answer 1

The correct option is D

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

At x=0,the value of the given function takes the form $\frac{0}{0}$

applying L'Hospital Rule

Differentiate Numerator and Denominator

$\underset{x\to 0}{lim}\frac{\frac{d}{dx}tan2x-\frac{d}{dx}x}{\frac{d}{dx}\left(3x\right)-\frac{d}{dx}sinx}$

= $\underset{x\to 0}{lim}\frac{se{c}^{2}2x\frac{d}{dx}\left(2x\right)-1}{3-cosx}$

= $\underset{x\to 0}{lim}\frac{2se{c}^{2}2x-1}{3-cosx}$

Putting x=0

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