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प्रश्न
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \cos \left( \frac{3 + x + 3 - x}{2} \right) \sin \left( \frac{3 + x - 3 + x}{2} \right)}{x} \right] \left\{ \because \sin C - \sin D = 2 \cos \left( \frac{C + D}{2} \right) \sin \left( \frac{C - D}{2} \right) \right\}\]
\[ = \lim_{x \to 0} \left[ \frac{2 \cos 3 \sin x}{x} \right]\]
\[ = 2 \cos 3\]
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