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The given mathematical expression is:

\[
\lim_{x \to 0} \frac{1 - \cos(3x)}{10x}
\]

To solve this limit, we will use the following tools:

1. **Series Expansion**: Use the Taylor series expansion of \(\cos(x)\) around \(x = 0\). The series expansion of \(\cos(x)\) is:
   \[
   \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots
   \]
   Applying this to \(\cos(3x)\), we get:
   \[
   \cos(3x) = 1 - \frac{(3x)^2}{2} + O(x^4)
   \]
   Simplifying the second term:
   \[
   \cos(3x) = 1 - \frac{9x^2}{2} + O(x^4)
   \]

2. **Substitute in the limit**:
   \[
   1 - \cos(3x) = 1 - \left(1 - \frac{9x^2}{2} + O(x^4)\right) = \frac{9x^2}{2} + O(x^4)
   \]
   So, the limit becomes:
   \[
   \lim_{x \to 0} \frac{\frac{9x^2}{2} + O(x^4)}{10x}
   \]

3. **Simplify**:
   \[
   \lim_{x \to 0} \frac{9x^2/2}{10x} = \lim_{x \to 0} \frac{9x}{20} = 0
   \]

Thus, the value of the limit is:

\[
\boxed{0}
\]

Would you like further clarification on any step or a more detailed explanation?

Here are 5 related questions to consider:
1. What is the Taylor series for sine and how does it differ from cosine?
2. How can L'Hopital's Rule be applied to solve limits like this?
3. What are other common trigonometric limits involving \(\cos\) and \(\sin\)?
4. How does the small-angle approximation for cosine help in solving limits?
5. Can the limit be solved using an alternative method, like a substitution?

**Tip**: Taylor expansions are particularly useful in approximating complex functions near specific points, especially around \(x = 0\) (Maclaurin series).

Find the limit of (1 - cos(3x)) / 10x as x approaches 0 using Taylor series expansion for cosine. Step-by-step breakdown suitable for undergraduate calculus students.

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