1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{\cos 2x - \cos x}{1 - \cos x}$$. 2. **Recall relevant formulas and rules:** - Use the cosine double-angle identity: $$\cos 2x = 2\cos^2 x - 1$$. - For small angles, $$\cos x \approx 1 - \frac{x^2}{2}$$. - The limit involves an indeterminate form \(\frac{0}{0}\), so we can use series expansions or trigonometric identities. 3. **Rewrite numerator using the double-angle formula:** $$\cos 2x - \cos x = (2\cos^2 x - 1) - \cos x = 2\cos^2 x - \cos x - 1$$. 4. **Approximate cosine near zero:** $$\cos x \approx 1 - \frac{x^2}{2}$$ $$\cos^2 x \approx \left(1 - \frac{x^2}{2}\right)^2 = 1 - x^2 + \frac{x^4}{4}$$ (higher order terms negligible) 5. **Substitute approximations into numerator:** $$2\cos^2 x - \cos x - 1 \approx 2\left(1 - x^2\right) - \left(1 - \frac{x^2}{2}\right) - 1 = 2 - 2x^2 - 1 + \frac{x^2}{2} - 1 = -\frac{3x^2}{2}$$. 6. **Approximate denominator:** $$1 - \cos x \approx 1 - \left(1 - \frac{x^2}{2}\right) = \frac{x^2}{2}$$. 7. **Form the limit expression:** $$\lim_{x \to 0} \frac{-\frac{3x^2}{2}}{\frac{x^2}{2}} = \lim_{x \to 0} -3 = -3$$. 8. **Conclusion:** The limit is $$-3$$.