$\lim\limits_{x\to 0}\large\frac{\sin(\pi\cos^2x)}{x^2}$ equals

Question

$(a)\;-\pi\qquad(b)\;\pi\qquad(c)\;\pi/2\qquad(d)\;1$

Answer 1

$\lim\limits_{x\to 0}\large\frac{\sin(\pi\cos^2x)}{x^2}=$$\lim\limits_{x\to 0}\large\frac{\sin(\pi-\pi\sin^2x)}{x^2}$

$[\sin(\pi-\theta)=\sin\theta]$

$\Rightarrow \lim\limits_{x\to 0}\large\frac{\sin(\pi\sin^2x)}{\pi\sin^2x}\times \frac{\pi\sin^2x}{x^2}$

$\Rightarrow \pi$

Hence (b) is the correct answer.