Discussion
\(\lim_{\theta \to 0} \frac{\theta\,sin2\theta}{1 - cos2\theta}\)
Answer: A
The limit is 1.
Use \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(1 - \cos 2\theta = 2\sin^{2}\theta\).
\[ \begin{aligned} \frac{\theta\cdot 2\sin\theta\cos\theta}{2\sin^{2}\theta} &= \frac{\theta}{\sin\theta}\cdot\cos\theta \\ &\to 1 \cdot 1 = 1 \end{aligned} \]
\(\frac{1}{2}\) comes from expanding \(1 - \cos 2\theta\) as \(2\theta^{2}\) correctly but forgetting the 2 in \(\sin 2\theta \approx 2\theta\). Both numerator and denominator are positive for small \(\theta\), so negative answers are impossible, and the limit is not 0 because both behave like \(2\theta^{2}\).
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