Question

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

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

Answer 1

Answer: (D)

$2$

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

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