1. **State the problem:** We need to find the limit $$\lim_{\theta \to 0} \frac{1 - \cos(\theta)}{2 \sin^2(\theta)}$$. 2. **Recall important formulas and rules:** - Use the trigonometric identity $$1 - \cos(\theta) = 2 \sin^2\left(\frac{\theta}{2}\right)$$. - As $$\theta \to 0$$, $$\sin(\theta) \approx \theta$$ and $$\sin\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}$$. 3. **Rewrite the expression using the identity:** $$\frac{1 - \cos(\theta)}{2 \sin^2(\theta)} = \frac{2 \sin^2\left(\frac{\theta}{2}\right)}{2 \sin^2(\theta)} = \frac{\sin^2\left(\frac{\theta}{2}\right)}{\sin^2(\theta)}$$. 4. **Apply the small angle approximations:** $$\sin\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}, \quad \sin(\theta) \approx \theta$$. 5. **Substitute approximations:** $$\frac{\sin^2\left(\frac{\theta}{2}\right)}{\sin^2(\theta)} \approx \frac{\left(\frac{\theta}{2}\right)^2}{\theta^2} = \frac{\frac{\theta^2}{4}}{\theta^2} = \frac{1}{4}$$. 6. **Conclusion:** Therefore, $$\lim_{\theta \to 0} \frac{1 - \cos(\theta)}{2 \sin^2(\theta)} = \frac{1}{4}$$. **Final answer:** C $$\frac{1}{4}$$.