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

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

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$\lim\limits_{x\to 0}\large\frac{\sin(\pi\cos^2x)}{x^2}\big(\large\frac{0}{0}\big)$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\cos(\pi\cos^2x)\pi(-\sin 2x)}{2x}$
$\Rightarrow (\lim\limits_{x\to 0}\large\frac{-\sin 2x}{2x}\big)$$\pi\lim\limits_{x\to 0}\cos(\pi\cos^2x)$
$\Rightarrow (-1)\pi(-1)$
$\Rightarrow \pi$
Hence (b) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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