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प्रश्न
\[\lim_{x \to 0} \frac{\sin x^2 \left( 1 - \cos x^2 \right)}{x^6}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\sin x^2 \left( 1 - \cos x^2 \right)}{x^6} \right]\]
\[= \lim_{x \to 0} \left[ \frac{\sin x^2 \times 2 \sin^2 \left( \frac{x^2}{2} \right)}{x^6} \right] \left[ \because 1 - \cos A = 2 \sin^2 \left( \frac{A}{2} \right) \right]\]
\[ = 2 \lim_{x \to 0} \left[ \frac{\sin x^2}{x^2} \times \frac{\sin \left( \frac{x^2}{2} \right)}{2 \times \frac{x^2}{2}} \times \frac{\sin \left( \frac{x^2}{2} \right)}{2 \times \frac{x^2}{2}} \right]\]
\[ = \frac{2}{2 \times 2}\]
\[ = \frac{1}{2}\]
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