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प्रश्न
\[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\tan \left( 2x \right) - \sin \left( 2x \right)}{x^3} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\frac{\sin 2x}{\cos 2x} - \sin 2x}{x^3} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin 2x \left\{ 1 - \cos 2x \right\}}{coS 2x \times x^3} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin 2x \times 2 \sin^2 x}{\cos 2x \times x^3} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin 2x}{2x} \times \frac{2}{\cos 2x} \times 2 \left( \frac{\sin x}{x} \right)^2 \right]\]
\[ = \frac{2 \times 2}{\cos 0}\]
\[ = 4\]
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