JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 9)
The value of the limit
$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ is equal to :
$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ is equal to :
0
$$-$$$${1 \over 2}$$
$${1 \over 4}$$
$$-$$$${1 \over 4}$$
Explanation
Given,
$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$
$$ = \mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi - \pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ $$ \therefore $$ $$\left( {{{\cos }^2}\theta = 1 - {{\sin }^2}\theta } \right)$$
$$ = \mathop {\lim }\limits_{\theta \to 0} {{ - \tan (\pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ $$ \therefore $$ $$(\tan (\pi - \theta ) = - \tan \theta )$$
$$ = \mathop {\lim }\limits_{\theta \to 0} {{{{ - \tan (\pi {{\sin }^2}\theta )} \over {\pi {{\sin }^2}\theta }}} \over {{{\sin (2\pi {{\sin }^2}\theta )} \over {2\pi {{\sin }^2}\theta }} \times 2}}\left( \matrix{ As\,\theta \to 0 \hfill \cr then \,{\sin ^2}\theta \to 0 \hfill \cr} \right)$$
$$ = -{1 \over 2}.$$ $$ \because $$ $$\left( \matrix{ \mathop {\lim }\limits_{\theta \to 0} {{\tan \theta } \over \theta } \to 1 \hfill \cr \& \,\mathop {\lim }\limits_{\theta \to 0} {{\sin \theta } \over \theta } = 1 \hfill \cr} \right)$$
$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$
$$ = \mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi - \pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ $$ \therefore $$ $$\left( {{{\cos }^2}\theta = 1 - {{\sin }^2}\theta } \right)$$
$$ = \mathop {\lim }\limits_{\theta \to 0} {{ - \tan (\pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ $$ \therefore $$ $$(\tan (\pi - \theta ) = - \tan \theta )$$
$$ = \mathop {\lim }\limits_{\theta \to 0} {{{{ - \tan (\pi {{\sin }^2}\theta )} \over {\pi {{\sin }^2}\theta }}} \over {{{\sin (2\pi {{\sin }^2}\theta )} \over {2\pi {{\sin }^2}\theta }} \times 2}}\left( \matrix{ As\,\theta \to 0 \hfill \cr then \,{\sin ^2}\theta \to 0 \hfill \cr} \right)$$
$$ = -{1 \over 2}.$$ $$ \because $$ $$\left( \matrix{ \mathop {\lim }\limits_{\theta \to 0} {{\tan \theta } \over \theta } \to 1 \hfill \cr \& \,\mathop {\lim }\limits_{\theta \to 0} {{\sin \theta } \over \theta } = 1 \hfill \cr} \right)$$
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