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 $2$

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