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$\lim\limits_{x\to 0}\large\frac{\cos(\sin x)-1}{x^2}$=

$(a)\;1\qquad(b)\;-1\qquad(c)\;\large\frac{1}{2}$$\qquad(d)\;-\large\frac{1}{2}$

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$\lim\limits_{x\to 0}\large\frac{\cos(\sin x)-1}{x^2}\qquad\big(\large\frac{0}{0}\big)$ form
$\Rightarrow \lim\limits_{x\to 0}\large\frac{-\sin(\sin x)\cos x}{2x}$
$\Rightarrow \large\frac{1}{2}$$\lim\limits_{x\to 0}\large\frac{\sin(\sin x)}{\sin x}.\big(\large\frac{\sin x}{x}\big)$$.\cos x$
$\Rightarrow \large\frac{-1}{2}$
Hence (d) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 

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