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If $f(x)$ is the integral function of the function $\large\frac{2\sin x-\sin 2x}{x^3}\qquad$$ x\neq 0$ then $\lim\limits_{x\to 0} f'(x)$ is

$(a)\;0\qquad(b)\;-1\qquad(c)\;1\qquad(d)\;None\;of\;these$

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Since $f(x)$ is integral function of $\large\frac{2\sin x-\sin 2x}{x^3}$
$\therefore f'(x)=\large\frac{2\sin x-\sin 2x}{x^3}$
$\Rightarrow \lim\limits_{x\to 0}f'(x)=\lim\limits_{x\to 0}\large\frac{2\sin x-\sin 2x}{x^3}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{2\sin x}{x}\big(\large\frac{1-\cos x}{x^2}\big)$
$\Rightarrow 2\times \large\frac{1}{2}$
$\Rightarrow 1$
Hence (c) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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