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}
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