1. **Problem Statement:** Evaluate the limit $$\lim_{\theta \to 0} \frac{\sin^2 \theta}{1 - \cos \theta}$$. 2. **Use Trigonometric Identities:** Recall the Pythagorean identity and the double-angle formula: $$1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right)$$ 3. **Rewrite the expression:** Substitute the denominator: $$\frac{\sin^2 \theta}{1 - \cos \theta} = \frac{\sin^2 \theta}{2 \sin^2 \left( \frac{\theta}{2} \right)}$$ 4. **Rewrite the numerator using double angle:** Using the identity $$\sin \theta = 2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)$$, then $$\sin^2 \theta = \left( 2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right) \right)^2 = 4 \sin^2 \left( \frac{\theta}{2} \right) \cos^2 \left( \frac{\theta}{2} \right)$$ 5. **Substitute numerator back:** $$\frac{4 \sin^2 \left( \frac{\theta}{2} \right) \cos^2 \left( \frac{\theta}{2} \right)}{2 \sin^2 \left( \frac{\theta}{2} \right)}$$ 6. **Simplify by cancelling $$\sin^2 \left( \frac{\theta}{2} \right)$$:** $$2 \cos^2 \left( \frac{\theta}{2} \right)$$ 7. **Evaluate the limit as $$\theta \to 0$$:** Since $$\cos(0) = 1$$, $$\lim_{\theta \to 0} 2 \cos^2 \left( \frac{\theta}{2} \right) = 2 \times 1^2 = 2$$ **Final answer:** $$\boxed{2}$$