JEE PYQ: Limits Question 4
Question 4 - 2021 (17 Mar Shift 2)
The value of the limit $\lim_{\theta \to 0} \frac{\tan(\pi\cos^2\theta)}{\sin(2\pi\sin^2\theta)}$ is equal to:
(1) $-\frac{1}{2}$
(2) $-\frac{1}{4}$
(3) $0$
(4) $\frac{1}{4}$
Show Answer
Answer: (1)
Solution
$\lim_{\theta\to 0} \frac{\tan(\pi(1-\sin^2\theta))}{\sin(2\pi\sin^2\theta)} = \lim_{\theta\to 0} \frac{-\tan(\pi\sin^2\theta)}{\sin(2\pi\sin^2\theta)}$. Using $\frac{\tan(\pi\sin^2\theta)}{\pi\sin^2\theta} \cdot \frac{2\pi\sin^2\theta}{\sin(2\pi\sin^2\theta)} \times \frac{1}{2} = -\frac{1}{2}$.
BETA
