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प्रश्न
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{1 - \cos \left( 5x \right)}{1 - \cos \left( 6x \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin^2 \left( \frac{5x}{2} \right)}{2 \sin^2 3x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin^2 \left( \frac{5x}{2} \right)}{\left( \frac{5x}{2} \right)^2} \times \frac{\left( \frac{5x}{2} \right)^2}{\frac{\sin^2 3x}{\left( 3x \right)^2} \times \left( 3x \right)^2} \right]\]
\[ = \lim_{x \to 0} \left[ \left\{ \frac{\sin \left( \frac{5x}{2} \right)}{\frac{5x}{2}} \right\}^2 \frac{\frac{25}{4} x^2}{\left\{ \frac{\sin \left( 3x \right)}{3x} \right\}^2 \times 9 x^2} \right]\]
\[ = \frac{25}{9 \times 4}\]
\[ = \frac{25}{36}\]
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