Step-by-Step Solution
Key Concept: Use Taylor: $\tan 2x \approx 2x + \frac{8x^3}{3}$, $\sin x \approx x - \frac{x^3}{6}$. Numerator $\approx x$, denominator $\approx 2x$. Limit $= \frac{x}{2x} = \frac{1}{2}$.
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Correct Answer: (1)