The value of the limit _ θ → 0 (π ^ 2 θ) sin (2π ^ 2 θ) is equal to...

We have to find $l = \lim_{\theta \to 0}\dfrac{\tan \left(\pi\cos^{2}\theta\right)}{\sin \left(2\pi\sin^{2}\theta\right)}$

Using $\sin^{2}\theta + \cos^{2}\theta = 1,$ we get

$⇒ l = \lim_{\theta \to 0}\dfrac{\tan \left(\pi\left(1 - \sin^{2}\theta\right)\right)}{\sin \left(2\pi\sin^{2}\theta\right)}$

$⇒ l = \lim_{\theta \to 0}\dfrac{\tan \left(\pi - \pi\sin^{2}\theta\right)}{\sin \left(2\pi\sin^{2}\theta\right)}$

Using $\tan \left(\pi - \theta\right) = - tan\theta,$ we get

$⇒ l = \lim_{\theta \to 0}\dfrac{- \tan \left(\pi\sin^{2}\theta\right)}{\sin \left(2\pi\sin^{2}\theta\right)}$

$⇒ l = \lim_{\theta \to 0} - \left(\dfrac{\tan \left(\pi\sin^{2}\theta\right)}{\pi\sin^{2}\theta}\right)\left(\dfrac{2\pi\sin^{2}\theta}{\sin \left(2\pi\sin^{2}\theta\right)}\right) \times \dfrac{1}{2}$

Using, $\lim_{x \to 0}\left(\dfrac{\sin x}{x}\right) = 1 \text{ and } \lim_{x \to 0}\left(\dfrac{\tan x}{x}\right) = 1,$ we get

$l = - \dfrac{1}{2} .$