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प्रश्न
\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{1 - \cos 2x}{\cos 2x - \cos 8x} \right]\]
\[= \lim_{x \to 0} \left[ \frac{2 \sin^2 x}{2 \sin\left( \frac{2x + 8x}{2} \right)\sin\left( \frac{8x - 2x}{2} \right)} \right] \left[ \because cosC - cosD = 2\sin\left( \frac{C + D}{2} \right)\sin\left( \frac{D - C}{2} \right) \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2\sin x \times \sin x}{2\sin 5x \times \sin 3x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin x \times \sin x}{\sin 5x \times \sin 3x} \right] \left[ \because \sin\left( - \theta \right) = - \sin\theta \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\frac{\sin x}{x} \times x \times \frac{\sin x}{x} \times x}{\frac{\sin 5x}{5x} \times 5x \times \frac{\sin 3x \times 3x}{3x}} \right]\]
\[ = \frac{1}{15}\]
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