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प्रश्न
Evaluate the following limits:
\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\]
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उत्तर
\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\]
\[ = \lim_{x \to 0} \frac{2\sin x - 2\sin x\cos x}{x^3}\]
\[ = \lim_{x \to 0} \frac{2\sin x\left( 1 - \cos x \right)}{x^3}\]
\[ = \lim_{x \to 0} \frac{2\sin x \times 2 \sin^2 \frac{x}{2}}{x^3}\]
\[ = \lim_{x \to 0} \frac{\sin x}{x} \times \left( \lim_{x \to 0} \frac{\sin\frac{x}{2}}{\frac{x}{2}} \right)^2 \]
\[ = 1 \times 1 \left( \lim_\theta \to 0 \frac{\sin\theta}{\theta} = 1 \right)\]
\[ = 1\]
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