To solve the limit \(\lim_{x \to 0} \frac{1 - \cos(2x)}{5x}\), we can use L'Hôpital's Rule because the expression is of the indeterminate form \(\frac{0}{0}\).

### Step 1: Confirm Indeterminate Form
As \(x \to 0\):
- The numerator \(1 - \cos(2x)\) approaches \(1 - \cos(0) = 0\).
- The denominator \(5x\) approaches \(0\).

So, we have a \(\frac{0}{0}\) form.

### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if a limit is of the form \(\frac{0}{0}\), we can take the derivatives of the numerator and the denominator:

\[
\lim_{x \to 0} \frac{1 - \cos(2x)}{5x} = \lim_{x \to 0} \frac{d}{dx}\left(1 - \cos(2x)\right) \bigg/ \frac{d}{dx}(5x)
\]

### Step 3: Differentiate the Numerator and Denominator
- The derivative of the numerator \(1 - \cos(2x)\) with respect to \(x\) is:
  \[
  \frac{d}{dx}\left(1 - \cos(2x)\right) = 2\sin(2x)
  \]
- The derivative of the denominator \(5x\) with respect to \(x\) is:
  \[
  \frac{d}{dx}(5x) = 5
  \]

### Step 4: Take the Limit
Now, we substitute back into the limit:

\[
\lim_{x \to 0} \frac{2\sin(2x)}{5}
\]

As \(x \to 0\), \(\sin(2x)\) approaches \(0\). Therefore, the limit simplifies to:

\[
\lim_{x \to 0} \frac{2\sin(2x)}{5} = \frac{2 \cdot 0}{5} = 0
\]

Thus, the limit is \(\boxed{0}\).

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Would you like more details or have any questions?

Here are some related questions you might consider:
1. How does L'Hôpital's Rule work for limits that are not in the \(\frac{0}{0}\) form?
2. What are some alternative methods to evaluate limits without L'Hôpital's Rule?
3. How does the small-angle approximation for sine and cosine functions help in limits?
4. How can Taylor series be used to approximate functions near certain points?
5. What is the geometric interpretation of this limit?

**Tip:** When you encounter a limit problem, first check if the function has an indeterminate form before applying L'Hôpital's Rule or other techniques.

Learn how to evaluate the limit \( \lim_{x \to 0} \frac{1 - \cos(2x)}{5x} \) using L'Hôpital's Rule. This problem involves trigonometric functions and provides a step-by-step solution. Suitable for advanced high school students and calculus learners.

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