1. The problem asks to find the limit as $x$ approaches 0 of $\sin^{2}(2x)$.\n\n2. We recognize that $\sin^{2}(2x)$ means $(\sin(2x))^2$.\n\n3. Using the property that $\sin(\theta) \approx \theta$ when $\theta$ approaches 0, we substitute $\theta = 2x$. So, $\sin(2x) \approx 2x$ as $x \to 0$.\n\n4. Therefore, $\sin^{2}(2x) \approx (2x)^2 = 4x^2$.\n\n5. Taking the limit as $x \to 0$, we have $\lim_{x \to 0} 4x^2 = 4 \times 0^2 = 0$.\n\n6. Hence, $\boxed{0}$ is the value of the limit.