Question

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

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

Answer 1

Answer: (A)

$\underset{\theta \to 0}{\lim }\frac{\tan \left(\pi \left(1-{\sin }^2\theta \right)\right)}{\sin \left(2\pi {\sin }^2\theta \right)}$
$=\underset{\theta \to 0}{\lim }\frac{-\tan \left(\pi {\sin }^2\theta \right)}{\sin \left(2\pi {\sin }^2\theta \right)}$
$=\underset{\theta \to 0}{\lim }-\left(\frac{\tan \left(\pi {\sin }^2\theta \right)}{\pi {\sin }^2\theta }\right)\left(\frac{2\pi {\sin }^2\theta }{\sin \left(2\pi {\sin }^2\theta \right)}\right)\times \frac{1}{2}$
$=\frac{-1}{2}$