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प्रश्न
\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin 2x \times - 2 \sin\left( \frac{3x + x}{2} \right) \sin\left( \frac{3x - x}{2} \right)}{x^3} \right]\]
\[ = - 2 \lim_{x \to 0} \left[ \frac{\sin 2x}{2x} \times \frac{\sin 2x}{2x} \times \frac{\sin x}{x} \right] \times 2 \times 2\]
\[ = - 8\]
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