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प्रश्न
\[\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\sin 5x - \sin 3x}{\sin x} \right]\]
\[= \lim_{x \to 0} \left[ \frac{2\cos\left( \frac{5x + 3x}{2} \right)\sin\left( \frac{5x - 3x}{2} \right)}{\sin x} \right] \left[ \because sinC - sinD = 2\cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right) \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2\cos4x . \sin x}{\sin x} \right]\]
\[ = 2 \cos0\]
\[ = 2\]
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