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प्रश्न
\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{1 - \cos \left( 4x \right)}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin^2 2x}{x^2} \right]\]
\[ = \lim_{x \to 0} 2\left[ \frac{\sin 2x}{x} \times \frac{\sin 2x}{x} \right]\]
\[ = \lim_{x \to 0} \left[ 2 \times \frac{\sin 2x}{2x} \times \frac{\sin2x}{2x} \times 4 \right]\]
\[ = 2 \times 1 \times 1 \times 4\]
\[ = 8\]
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