We are tasked with evaluating the limit:

$\lim_{x \to 0} \frac{\cos(x) - \cos(3x)}{1 - \cos^2(x)}.$

Step-by-Step Solution

1. Simplify the Denominator:
   Notice that $1 - \cos^2(x)$ is a well-known trigonometric identity: $1 - \cos^2(x) = \sin^2(x).$ So, the expression becomes: $\frac{\cos(x) - \cos(3x)}{\sin^2(x)}.$

2. Expand the Numerator:
   We can use the sum-to-product formula for the difference of cosines: $\cos(A) - \cos(B) = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right).$ Applying this to $\cos(x) - \cos(3x)$, we get:

   = -2 \sin(2x) \sin(-x) = 2 \sin(2x) \sin(x).$$ So, the expression becomes: $$\frac{2 \sin(2x) \sin(x)}{\sin^2(x)}.$$

3. Simplify the Expression: Cancel one $\sin(x)$ from the numerator and denominator: $\frac{2 \sin(2x)}{\sin(x)}.$ Using the identity $\sin(2x) = 2 \sin(x) \cos(x)$, we substitute: $\frac{2 \cdot 2 \sin(x) \cos(x)}{\sin(x)} = 4 \cos(x).$

4. Evaluate the Limit:
   Now, evaluate the limit as $x \to 0$: $\lim_{x \to 0} 4 \cos(x) = 4 \cos(0) = 4 \cdot 1 = 4.$

Final Answer:

$\lim_{x \to 0} \frac{\cos(x) - \cos(3x)}{1 - \cos^2(x)} = 4.$

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. What are the sum-to-product formulas in trigonometry, and how are they derived?
2. How would you compute the limit if the numerator was $\sin(x) - \sin(3x)$ instead?
3. Why is $\sin^2(x)$ used frequently in limit problems, especially around $x = 0$?
4. How would you approach limits that involve higher powers of $\sin(x)$ or $\cos(x)$?
5. How can L'Hopital's Rule be applied to similar trigonometric limits?

Tip: When dealing with trigonometric limits, always check for identities like small angle approximations or product-to-sum formulas.