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प्रश्न
\[\lim_{x \to 0} \frac{\sin \left( a + x \right) + \sin \left( a - x \right) - 2 \sin a}{x \sin x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\sin \left( a + x \right) + \sin \left( a - x \right) - 2 \sin a}{x \sin x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin \left( \frac{a + x + a - x}{2} \right) \cos \left( \frac{a + x - a + x}{2} \right) - 2 \sin a}{x \sin x} \right] \left\{ \because \sin C + \sin D = 2 \sin \left( \frac{C + D}{2} \right)\cos \left( \frac{C - D}{2} \right) \right\}\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin a \cos x - 2 \sin a}{x \sin x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin a \left( \cos x - 1 \right)}{x \sin x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin a \left( 1 - 2 \sin^2 \frac{x}{2} - 1 \right)}{x \sin x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin a \left( - 2 \sin^2 \frac{x}{2} \right)}{x \sin x} \right]\]
\[ = - 4 \sin a \lim_{x \to 0} \left[ \frac{1}{\frac{x \sin x}{x^2}} \times \frac{\sin \left( \frac{x}{2} \right)}{\frac{x}{2}} \times \frac{\sin\left( \frac{x}{2} \right)}{\frac{x}{2}} \times \frac{1}{4} \right]\]
\[ = - 4 \sin a \times \frac{1}{1} \times 1 \times 1 \times \frac{1}{4}\]
\[ = - \sin a\]
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