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$\lim\limits_{x\to 0}\large\frac{(\cos x)^{1/2}-(\cos x)^{1/3}}{\sin^2x}$ is

$(a)\;1/6\qquad(b)\;-1/12\qquad(c)\;2/3\qquad(d)\;1/3$

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Given limit (0/0)
$\Rightarrow \lim\limits_{x\to 0}\large\frac{-\Large\frac{1}{2}\normalsize (\cos x)^{\Large -1/2}\sin x+\Large\frac{1}{3}\normalsize(\cos x)^{\Large -2/3}\sin x}{2\sin x\cos x}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{-1/2(\cos x)^{-1/2}+1/3(\cos x)^{-2/3}}{2\cos x}$
$\Rightarrow \large\frac{1}{2}\bigg[\frac{-1}{2}+\frac{1}{3}\bigg]$
$\Rightarrow -\large\frac{1}{12}$
Hence (b) is the correct answer.
answered Dec 23, 2013 by sreemathi.v
 

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