Evaluation of the Form $$1^\infty$$

The limit is of the form $$1^\infty$$ as $$x \to 0$$.

1. Apply the standard formula:
$$L = e^{\lim_{x \to 0} \frac{1}{x^2} \left(2 - \cos x \sqrt{\cos 2x} - 1\right)} = e^{\lim_{x \to 0} \frac{1 - \cos x \sqrt{\cos 2x}}{x^2}}$$

2. Use Taylor series expansion for the trigonometric functions:
$$\cos x = 1 - \frac{x^2}{2} + \dots$$
$$\cos 2x = 1 - \frac{(2x)^2}{2} = 1 - 2x^2 + \dots$$
Using the binomial expansion for the square root:
$$\sqrt{\cos 2x} = (1 - 2x^2)^{1/2} = 1 - \frac{1}{2}(2x^2) = 1 - x^2 + \dots$$

3. Multiply the expansions in the numerator:
$$\cos x \sqrt{\cos 2x} \approx \left(1 - \frac{x^2}{2}\right)(1 - x^2) \approx 1 - x^2 - \frac{x^2}{2} = 1 - \frac{3x^2}{2}$$

4. Evaluate the limit inside the exponent:
$$1 - \cos x \sqrt{\cos 2x} \approx 1 - \left(1 - \frac{3x^2}{2}\right) = \frac{3x^2}{2}$$
Substituting this back gives:
$$L = e^{\lim_{x \to 0} \frac{3x^2 / 2}{x^2}} = e^{3/2}$$