Question

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

• A. 4
• B. 3
• C. 2
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

We have, $\underset{x\to 0}{\lim }\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}=\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x(3+\cos x)}{x\times \frac{\tan 4x}{4x}\times 4x}$ $=\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x}{x^2}\times \underset{x\to 0}{\lim }\frac{(3+\cos x)}{4}\times \frac{1}{\underset{x\to 0}{\lim }\frac{\tan 4x}{4x}}$ $=2\times \frac{4}{4}\times 1\left[\because \underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1\text{and }\underset{\theta \to 0}{\lim }\frac{\tan \theta }{\theta }=1\right]$ $=2$