Question

The value of the limit $\underset{\theta \to 0}{\lim }\frac{\left[\tan \left({\mathrm{\pi cos}}^2\left(\mathrm{\theta }\right)\right)\right]}{\left[\sin (2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right))\right]}$ is equal to

• A. $-\frac{1}{2}$
• B. $-\frac{1}{4}$
• C. $0$
• D. $\frac{1}{4}$

Answer 1

The correct option is A

$-\frac{1}{2}$

Explanation for the correct option

Given function is $\underset{\theta \to 0}{\lim }\frac{\left[\tan \left({\mathrm{\pi cos}}^2\left(\mathrm{\theta }\right)\right)\right]}{\left[\sin (2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right))\right]}$

$$\begin{gathered} =\underset{\theta \to 0}{\lim }\frac{\left[\tan \left(\pi \right(1-{\sin }^2\left(\mathrm{\theta }\right))\right]}{\left[\sin (2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right))\right]}[\because {\sin }^2\left(\mathrm{\theta }\right)+{\cos }^2\left(\mathrm{\theta }\right)=1] \\ =\underset{\theta \to 0}{\lim }\frac{-\left[\tan \left(\pi {\sin }^2\left(\mathrm{\theta }\right)\right)\right]}{\left[\sin (2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right))\right]} \end{gathered}$$

Now, multiply both the numerator and denominator by $2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right)$, we get,

$$\begin{gathered} =\underset{\theta \to 0}{\lim }\frac{-\left[\tan \left(\pi {\sin }^2\left(\mathrm{\theta }\right)\right)\right]}{\left[\sin (2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right))\right]}\times \frac{2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right)}{2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right)} \\ =\underset{\theta \to 0}{\lim }-\frac{\left[\tan \left(\pi {\sin }^2\left(\mathrm{\theta }\right)\right)\right]}{\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right)}\times \frac{2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right)}{2\left[\sin (2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right))\right]} \\ =\underset{\theta \to 0}{\lim }-\frac{\left[\tan \left(\pi {\sin }^2\left(\mathrm{\theta }\right)\right)\right]}{\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right)}\times \frac{2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right)}{\left[\sin (2\mathrm{\pi }{\sin }^2\left(\mathrm{\theta }\right))\right]}\times \frac{1}{2} \\ =-\frac{1}{2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}[\because \underset{\mathrm{\theta }\to 0}{\lim }\frac{\tan \left(\mathrm{\theta }\right)}{\mathrm{\theta }}=1,\underset{\mathrm{\theta }\to 0}{\lim }\frac{\sin \left(\mathrm{\theta }\right)}{\mathrm{\theta }}=1]\mathbf{} \end{gathered}$$

Hence, option(A) i.e. $-\frac{1}{2}$ is the correct answer.