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प्रश्न
\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\]
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उत्तर
\[= \lim_{x \to 0} \left[ \frac{2 \cos\left( \frac{3x + x}{2} \right) \sin\left( \frac{3x - x}{2} \right)}{\sin x} \right] \left[ \because \sin C - \sin D = 2\cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right) \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \cos2x . \sin x}{\sin x} \right]\]
\[ = 2\cos0\]
\[ = 2\]
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