1. State the problem: We want to find the limit $$\lim_{x \to 0} \frac{\sin^{2}(2x)}{1 - \cos(2x)}.$$\n\n2. Use trigonometric identities to simplify: Recall the identity $$1 - \cos(2x) = 2\sin^{2}(x).$$\n\n3. Substitute into the limit expression:\n$$\lim_{x \to 0} \frac{\sin^{2}(2x)}{2\sin^{2}(x)}.$$\n\n4. Rewrite using the double angle for sine: $$\sin(2x) = 2\sin(x)\cos(x),$$ so:\n$$\sin^{2}(2x) = (2\sin(x)\cos(x))^{2} = 4\sin^{2}(x)\cos^{2}(x).$$\n\n5. Substitute back into the limit:\n$$\lim_{x \to 0} \frac{4\sin^{2}(x)\cos^{2}(x)}{2\sin^{2}(x)} = \lim_{x \to 0} 2\cos^{2}(x).$$\n\n6. As $x \to 0$, $\cos(x) \to 1$, so:\n$$2\cos^{2}(0) = 2 \times 1^{2} = 2.$$\n\nTherefore, the limit is \boxed{2}.