Evaluate: $$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}$$

Step 1: Apply L'Hôpital's Rule (since it's 0/0):

First derivative: $$\frac{e^x + e^{-x} - 2}{\sin x}$$

Still 0/0 → Apply L'Hôpital's Rule again: $$\frac{e^x - e^{-x}}{\cos x}$$

Now, $$\lim_{x \to 0} \frac{1 - 1}{1} = 0$$

Final Answer: $$\boxed{0}$$