JEE Main 2021 (Online) 17th March Evening ShiftQuestion 1
The value of the limit $\lim_{\theta\to0}\frac{\tan\left(\pi\cos^{2}\theta\right)}{\sin\left(2\pi\sin^{2}\theta\right)}$ is equal to :
Limits & Continuity
Solution
$\tan(\pi\cos^2\theta)=\tan(\pi-\pi\sin^2\theta)=-\tan(\pi\sin^2\theta)\sim-\pi\theta^2$. $\sin(2\pi\sin^2\theta)\sim2\pi\theta^2$. Limit $=-\frac12$: option B.