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

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

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$\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.
answered Dec 20, 2013 by sreemathi.v
 

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