Find \( \lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\sin ^{2}(2 \theta)}{1-\sin ^{2}(\theta)} \)
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To find the limit \[ \lim_{\theta \rightarrow \frac{\pi}{2}} \frac{\sin^{2}(2\theta)}{1 - \sin^{2}(\theta)}, \] we start by simplifying the expression. First, we know that \( 1 - \sin^{2}(\theta) = \cos^{2}(\theta) \). Thus, the limit can be rewritten as: \[ \lim_{\theta \rightarrow \frac{\pi}{2}} \frac{\sin^{2}(2\theta)}{\cos^{2}(\theta)}. \] Next, we substitute \( \sin(2\theta) \) with its identity: \[ \sin(2\theta) = 2\sin(\theta)\cos(\theta). \] So we have: \[ \sin^{2}(2\theta) = [2\sin(\theta)\cos(\theta)]^{2} = 4\sin^{2}(\theta)\cos^{2}(\theta). \] Substituting this back into our limit gives: \[ \frac{\sin^{2}(2\theta)}{\cos^{2}(\theta)} = \frac{4\sin^{2}(\theta)\cos^{2}(\theta)}{\cos^{2}(\theta)}. \] For \( \cos(\theta) \neq 0 \), this simplifies to: \[ 4\sin^{2}(\theta). \] Now, we take the limit of \( 4\sin^{2}(\theta) \) as \( \theta \) approaches \( \frac{\pi}{2} \): \[ \sin\left(\frac{\pi}{2}\right) = 1 \quad \text{therefore} \quad \sin^{2}\left(\frac{\pi}{2}\right) = 1. \] Thus, we calculate: \[ \lim_{\theta \rightarrow \frac{\pi}{2}} 4\sin^{2}(\theta) = 4 \cdot 1 = 4. \] Therefore, the value of the limit is \[ \boxed{4}. \]
