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$\lim\limits_{x\to 0}\large\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}$=


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$\lim\limits_{x\to 0}\large\frac{2\sin^2x\sin 5x}{x^2\sin 3x}$
$\Rightarrow \lim\limits_{x\to 0}2\big(\large\frac{\sin x}{x}\big)^2\frac{\sin 5x}{5x}.\frac{3x}{\sin 3x}.\frac{5}{3}$
$\Rightarrow 2(1)^2\times 1\times 1\times \large\frac{5}{3}$
$\Rightarrow \large\frac{10}{3}$
Hence (a) is the correct answer.
answered Dec 24, 2013 by sreemathi.v

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