Question

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

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

Answer 1

The correct option is D

$2$

Explanation for the correct option:

Finding the value of the given limit:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x} \\ =\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x(3+\cos x)}{x\times {\displaystyle \frac{\tan 4x}{4x}}\times 4x}\left[\because x\tan 4x=x\times \frac{\tan 4x}{4x}\times 4xand1-\cos 2x=2{\sin }^{2}x\right] \\ =\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x}{x^2}\times \underset{x\to 0}{\lim }\frac{3+\cos \left(x\right)}{4}\times \frac{1}{\underset{x\to 0}{\lim }\frac{\tan 4x}{4x}} \end{gathered}$$

As we know, $\underset{\mathbf{\theta }\mathbf{\to }\mathbf{0}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\left(\theta \right)}{\mathbf{\theta }}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{\&}\mathbf{}\underset{\mathbf{\theta }\mathbf{\to }\mathbf{0}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\left(\theta \right)}{\mathbf{\theta }}\mathbf{=}\mathbf{1}$

Applying the limits,

$$\begin{gathered} =2{\left(1\right)}^2\times \frac{3+1}{4}\times \frac{1}{1} \\ =2\times \frac{4}{4}\times 1 \\ =2\times 1\times 1 \\ =2 \end{gathered}$$

Therefore, the correct answer is option (D).