We are tasked with evaluating the limit:

\[
\lim_{x \to 0} \frac{\cos x - \cos 2x}{x^2}
\]

### Step-by-step solution:

1. **Apply L'Hopital's Rule**:
   Since both the numerator and the denominator approach 0 as \( x \to 0 \) (an indeterminate form of \( \frac{0}{0} \)), we can apply L'Hopital's Rule. This requires taking the derivative of both the numerator and the denominator.

2. **First derivative of the numerator**:
   The numerator is \( \cos x - \cos 2x \).
   - The derivative of \( \cos x \) is \( -\sin x \).
   - The derivative of \( \cos 2x \) is \( -2\sin 2x \) (using the chain rule).

   So, the derivative of the numerator is:
   \[
   \frac{d}{dx} [\cos x - \cos 2x] = -\sin x + 2\sin 2x
   \]

3. **First derivative of the denominator**:
   The denominator is \( x^2 \), and the derivative of \( x^2 \) is \( 2x \).

4. **New limit expression**:
   After applying L'Hopital's Rule once, we now evaluate the following limit:
   \[
   \lim_{x \to 0} \frac{-\sin x + 2\sin 2x}{2x}
   \]
   This is still an indeterminate form \( \frac{0}{0} \), so we apply L'Hopital's Rule again.

5. **Second derivative of the numerator**:
   The numerator is now \( -\sin x + 2\sin 2x \).
   - The derivative of \( -\sin x \) is \( -\cos x \).
   - The derivative of \( 2\sin 2x \) is \( 4\cos 2x \) (using the chain rule again).

   So, the second derivative of the numerator is:
   \[
   \frac{d}{dx} [-\sin x + 2\sin 2x] = -\cos x + 4\cos 2x
   \]

6. **Second derivative of the denominator**:
   The second derivative of \( 2x \) is just 2.

7. **Final limit expression**:
   After applying L'Hopital's Rule again, we now evaluate:
   \[
   \lim_{x \to 0} \frac{-\cos x + 4\cos 2x}{2}
   \]
   At \( x = 0 \):
   - \( \cos 0 = 1 \)
   - \( \cos 2(0) = \cos 0 = 1 \)

   Substituting these values into the limit:
   \[
   \frac{-1 + 4(1)}{2} = \frac{-1 + 4}{2} = \frac{3}{2}
   \]

### Conclusion:
The value of the limit is:
\[
\lim_{x \to 0} \frac{\cos x - \cos 2x}{x^2} = \frac{3}{2}
\]

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Explore the step-by-step process to evaluate the limit of (cos x - cos 2x) / x² as x approaches 0, utilizing L'Hopital's Rule. This solution covers derivatives, trigonometric functions, and advanced calculus techniques. Suitable for students in Grades 11-12 or early university level.

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