Question

$\underset{x\to 0}{lim}\frac{(1-\cos 2x)(3+\cos 3x)}{x\tan 4x}$ is equal to :

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

Answer 1

The correct option is D $2$

$\underset{x\to 0}{lim}\frac{(1-\cos 2x)(3+\cos 3x)}{x\tan 4x}=\underset{x\to 0}{lim}\frac{\left(2{\sin }^{2}x\right)(3+\cos 3x)}{x\tan 4x}=\underset{x\to 0}{lim}\left(\frac{2}{4}\right)\left(\frac{3+\cos 3x}{1}\right)\left(\frac{{\sin }^{2}x}{x^2}\right)\left(\frac{4x}{\tan 4x}\right)=\frac{1}{2}\cdot (3+1)=2$