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. $1$

Answer 1

The correct option is C $2$

$L{t}_{x\to 0}\frac{(1-\cos 2x)(3+\cos x)}{xtan4x}\Rightarrow L{t}_{x\to 0}\frac{2{\sin }^{2}x(3+\cos x)}{x^2}\times \frac{4x}{tan4x}\times \frac{1}{4}L{t}_{x\to 0}(\frac{\sin x}{x}{)}^2\times L{t}_{x\to 0}(\frac{4x}{tan4x})\times \frac{2\times (3+\cos x)}{4}\Rightarrow 1\times 1\times \frac{(3+1)}{4}\times 2\Rightarrow 2$