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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