Question

$\underset{\theta \to 0}{\lim }\left(\frac{4\theta (\tan \theta -2\theta \tan \theta )}{(1-\cos 2\theta )}\right)=$

• A. $\frac{1}{\sqrt{2}}$
• 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:

$$\begin{gathered} \underset{\theta \to 0}{\lim }\left(\frac{4\theta (\tan \theta -2\theta \tan \theta )}{(1-\cos 2\theta )}\right) \\ \mathrm{Multiply}\mathrm{and}\mathrm{Divide}\mathrm{by}{\theta }^{\mathit{2}} \\ =\underset{\theta \to 0}{\lim }\left(\frac{4\theta (\tan \theta -2\theta \tan \theta )}{2{\sin }^2\theta }\times \frac{{\theta }^2}{{\theta }^2}\right)\mathbf{}\left[\because 2{\sin }^2\theta =1-\cos 2\theta \right] \\ =2\underset{\theta \to 0}{\lim }\left(\frac{\theta (\tan \theta -2\theta \tan \theta )}{{\theta }^2}\times \frac{{\theta }^2}{{\sin }^2\theta }\right) \\ =2\underset{\theta \to 0}{\lim }\left(\frac{(\tan \theta -2\theta \tan \theta )}{\theta }\times {\left(\frac{\theta }{\sin \theta }\right)}^2\right) \end{gathered}$$

Applying the limits,

$$\begin{gathered} =2(1-2\times 0)\times {\left(1\right)}^2 \\ =2\left(1\right)\left(1\right) \\ =2 \end{gathered}$$

Therefore, the correct answer is option (D).