This problem can be solved by applying the L'Hopital rule twice:

 $$\lim_{x\to 0}\frac{(1-\cos{x})^2}{x^2-x^2\cos^2{x}}=\lim_{x\to 0}\frac{(1-\cos{x})^2}{x^2\sin^2{x}}=\left(\lim_{x\to 0}\frac{1-\cos{x}}{x\sin{x}}\right)^2=\left(\lim_{x\to 0}\frac{\sin{x}}{\sin{x}+x\cos{x}}\right)^2=\left(\lim_{x\to 0}\frac{\cos{x}}{\cos{x}+\cos{x}-x\sin{x}}\right)^2=\boxed{\frac{1}{4}}$$