1. **State the problem:** Find the limit $$\lim_{x \to 0} \left( \frac{1}{2 - 2\cos x} - \frac{1}{x^2} \right).$$ 2. **Recall important formulas and rules:** - Use the Taylor series expansion for cosine near 0: $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$$ - Simplify the expression by substituting the expansion and then combine terms. 3. **Substitute the expansion:** $$2 - 2\cos x = 2 - 2\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) = 2 - 2 + x^2 - \frac{x^4}{12} = x^2 - \frac{x^4}{12}.$$ 4. **Rewrite the original limit:** $$\lim_{x \to 0} \left( \frac{1}{x^2 - \frac{x^4}{12}} - \frac{1}{x^2} \right).$$ 5. **Find a common denominator and combine:** $$= \lim_{x \to 0} \left( \frac{1}{x^2\left(1 - \frac{x^2}{12}\right)} - \frac{1}{x^2} \right) = \lim_{x \to 0} \frac{1 - \left(1 - \frac{x^2}{12}\right)}{x^2\left(1 - \frac{x^2}{12}\right)} = \lim_{x \to 0} \frac{\frac{x^2}{12}}{x^2\left(1 - \frac{x^2}{12}\right)}.$$ 6. **Cancel $x^2$ terms:** $$= \lim_{x \to 0} \frac{\cancel{x^2} / 12}{\cancel{x^2} \left(1 - \frac{x^2}{12}\right)} = \lim_{x \to 0} \frac{1/12}{1 - \frac{x^2}{12}}.$$ 7. **Evaluate the limit as $x \to 0$:** $$= \frac{1}{12} \times \frac{1}{1 - 0} = \frac{1}{12}.$$ **Final answer:** $$\boxed{\frac{1}{12}}.$$