The limit problem provided is:

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

To evaluate this limit, we will use a known limit result:

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

Using this result, we can simplify the given expression:

$\lim_{x \to 0} \frac{1 - \cos(x)}{5x^2} = \frac{1}{5} \cdot \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{5} \cdot \frac{1}{2} = \frac{1}{10}$

Thus, the value of the limit is:

$\boxed{\frac{1}{10}}$

Would you like more details, or do you have any questions?

Here are 5 related questions to consider:

1. How do you apply L'Hopital's Rule to similar limit problems?
2. What is the Taylor series expansion for $\cos(x)$ and how can it help in solving limits?
3. How does the small-angle approximation for $\cos(x)$ help in simplifying limits?
4. What other limits are useful to memorize for trigonometric functions?
5. How would this limit change if the function involved $\sin(x)$ instead of $\cos(x)$?

Tip: The Taylor series expansion of a function can often simplify evaluating limits, especially for trigonometric functions.