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$\lim\limits_{x\to 0}\large\frac{\sin x-x}{x^3}$=

$(a)\;\large\frac{1}{3}$$\qquad(b)\;\large\frac{-1}{3}$$\qquad(c)\;\large\frac{1}{6}$$\qquad(d)\;-\large\frac{1}{6}$

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$\lim\limits_{x\to 0}\large\frac{\sin x-x}{x^3}\qquad\big(\large\frac{0}{0}\big)$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\cos x-1}{3x^2}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{-\sin x}{6x}$
$\Rightarrow -\large\frac{1}{6}$
Hence (d) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 

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