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

$(a)\;1\qquad(b)\;0\qquad(c)\;2\qquad(d)\;1/2$

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$f(x)=\int \large\frac{2\sin x-\sin 2x}{x^3}$$dx\qquad x\neq 0$
$\therefore f'(x)=\large\frac{2\sin x-\sin 2x}{x^3}$$\qquad x\neq0$
$\lim\limits_{x\to 0}f'(x)=\lim\limits_{x\to 0}\large\frac{2\sin x-\sin 2x}{x^3}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{2\sin x(1-\cos x)(1+\cos x)}{x^3(1+\cos x)}$
$\Rightarrow \lim\limits_{x\to 0}2\large\frac{\sin^3x}{x^3}.\frac{1}{1+\cos x}$
$\Rightarrow 2\times 1^3\times \large\frac{1}{2}$
$\Rightarrow 1$
Hence (a) is the correct answer.
answered Dec 31, 2013 by sreemathi.v
 

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