Mathematics - Limits Question with Solution | TestHub

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

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

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

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

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

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

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

Using, $\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)=1 \& \underset{x\to 0}{\lim }\left(\frac{\tan x}{x}\right)=1,$ we get

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