Step 1: Apply the identity $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$ to both numerator and denominator.

Step 2: For numerator: $\cos 2x - 1 = -(1 - \cos 2x) = -2 \sin^2 x$

Step 3: For denominator: $\cos x - 1 = -(1 - \cos x) = -2 \sin^2 \left(\frac{x}{2}\right)$

Step 4: Simplify: $\lim_{x \to 0} \frac{-2 \sin^2 x}{-2 \sin^2 \left(\frac{x}{2}\right)} = \lim_{x \to 0} \frac{\sin^2 x}{\sin^2 \left(\frac{x}{2}\right)}$

Step 5: Rewrite as: $\lim_{x \to 0} \left[\frac{\sin x}{x} \cdot \frac{x/2}{\sin \left(\frac{x}{2}\right)} \cdot 2\right]^2$

Step 6: Using $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{\sin \left(\frac{x}{2}\right)}{x/2} = 1$, we get: $(1 \times \frac{1}{1} \times 2)^2 = 4$