The given limit problem is:

$\lim_{{x \to 0}} \frac{1 - \cos(3x)}{\cos^2(5x) - 1}$

Step 1: Analyze the denominator

We know that $\cos^2(5x) - 1$ is a standard trigonometric identity. Using the identity $\cos^2(\theta) - 1 = -\sin^2(\theta)$, we rewrite the denominator:

$\cos^2(5x) - 1 = -\sin^2(5x)$

Thus, the limit becomes:

$\lim_{{x \to 0}} \frac{1 - \cos(3x)}{-\sin^2(5x)}$

Step 2: Use small-angle approximations

For small $x$, we use the standard approximations:

• $\cos(x) \approx 1 - \frac{x^2}{2}$
• $\sin(x) \approx x$

Using this for $\cos(3x)$ and $\sin(5x)$:

• $1 - \cos(3x) \approx 1 - \left(1 - \frac{(3x)^2}{2}\right) = \frac{9x^2}{2}$
• $\sin^2(5x) \approx (5x)^2 = 25x^2$

Substituting these into the limit expression:

$\lim_{{x \to 0}} \frac{\frac{9x^2}{2}}{-25x^2}$

Step 3: Simplify the expression

The $x^2$ terms cancel out, leaving:

$\lim_{{x \to 0}} \frac{9}{-50} = -\frac{9}{50}$

Thus, the value of the limit is:

$\boxed{-\frac{9}{50}}$

Let me know if you need more details or have any questions!

Here are 5 related questions to expand on this concept:

1. How do you apply L'Hopital's Rule to limits involving trigonometric functions?
2. What is the Taylor series expansion of $\cos(x)$ around $x = 0$?
3. Why does the small-angle approximation work for limits as $x \to 0$?
4. How does the limit change if the trigonometric functions had different coefficients for $x$?
5. What are some other useful trigonometric identities for solving limits?

Tip: When dealing with limits involving trigonometric functions as $x \to 0$, small-angle approximations are powerful tools to simplify the calculations.