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$\lim_{x \rightarrow 0}\frac{\sin(\pi\cos^{2}x)}{x^{2}}$ is equal to

$\frac{\pi}{2}$

B) 1

$\pi$

D) -

$\pi$

Option D

$\lim_{x \rightarrow 0}\frac{\sin(\pi\cos^{2}x)}{x^{2}}$

= $\lim_{x \rightarrow 0}\frac{\sin\pi(1-\sin^{2}x)}{x^{2}}$

= $\lim_{x \rightarrow 0}\frac{\sin(\pi-\pi\sin^{2}x)}{x^{2}}$

= $\lim_{x \rightarrow 0}\frac{\sin(\pi\sin^{2}x)}{x^{2}}$

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

$=\lim_{x \rightarrow0}\frac{\sin\pi\sin ^{2}x}{\pi\sin^{2}x}\times (\pi)\left[\frac{\sin^{2}x}{x^{2}}\right]$

= $\pi$ [ $\lim_{x \rightarrow0}\frac{\sin\theta}{\theta}=1$ ]

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