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

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

1 Answer

Given limit $\big(0/0\big)$
$\Rightarrow\lim\limits_{x\to 0}\large\frac{-(1/2)(\cos x)^{-1/2}\sin x+(1/3)(\cos x)^{-2/3}\sin x}{2\sin x\cos x}$
(L Hospital Rule)
$\Rightarrow \lim\limits_{x\to 0}\large\frac{(-1/2)(\cos x)^{-1/2}+1/3(\cos x)^{-2/3}}{2\cos x}$
$\Rightarrow \large\frac{1}{2}\big(-\large\frac{1}{2}+\frac{1}{3}\big)$
$\Rightarrow -\large\frac{1}{12}$
Hence (b) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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