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)

2

$\underset{x\to 0}{lim}\frac{(1-cos2x)(3+cosx)}{x\tan 4x}$

$=\underset{x\to 0}{lim}\frac{2{\sin }^{2}x(3+cosx)}{x\tan 4x}$

$=\underset{x\to 0}{lim}\frac{2{\sin }^{2}x(3+cosx)}{4x^2\frac{\tan 4x}{4x}}$

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

$=\frac{1^2(3+1)}{2.1}=2$