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To evaluate the limit

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

using L'Hôpital's Rule, we need to verify if this is an indeterminate form. As \( x \to 0 \), both the numerator \( \sin(5x) \) and the denominator \( x \) approach \( 0 \). Thus, we have a \( \frac{0}{0} \) indeterminate form, which allows us to apply L'Hôpital's Rule.

### Step-by-Step Solution:

1. **Differentiate the Numerator and Denominator:**
   - Differentiate the numerator: \( \frac{d}{dx}[\sin(5x)] = 5 \cos(5x) \).
   - Differentiate the denominator: \( \frac{d}{dx}[x] = 1 \).

2. **Apply L'Hôpital's Rule:**
   \[
   \lim_{{x \to 0}} \frac{\sin(5x)}{x} = \lim_{{x \to 0}} \frac{5 \cos(5x)}{1}.
   \]

3. **Evaluate the New Limit:**
   Substitute \( x = 0 \) into the expression:
   \[
   \lim_{{x \to 0}} 5 \cos(5x) = 5 \cos(0) = 5 \cdot 1 = 5.
   \]

### Final Answer:
\[
\lim_{{x \to 0}} \frac{\sin(5x)}{x} = 5.
\]

Would you like more details or have any questions?

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Here are some related questions for further exploration:

1. How does L'Hôpital's Rule apply to limits at infinity?
2. What is the limit of \( \frac{\sin(kx)}{x} \) as \( x \to 0 \) for a constant \( k \)?
3. Can L'Hôpital's Rule be used for higher-order indeterminate forms, and if so, how?
4. How would this limit change if the numerator were \( \sin(5x^2) \) instead?
5. What is the graphical interpretation of this limit result?

**Tip:** When you encounter a \( \frac{0}{0} \) form in a limit, L'Hôpital's Rule is a powerful tool to simplify the expression, especially when trigonometric functions are involved.

Use L'Hôpital's Rule to evaluate the limit \( \lim_{{x \to 0}} \frac{\sin(5x)}{x} \).

Learn how to evaluate the limit \( \lim_{{x \to 0}} \frac{\sin(5x)}{x} \) using L'Hôpital's Rule. This solution covers essential calculus concepts, including limits, indeterminate forms, and L'Hôpital's Rule application. Suitable for undergraduate calculus students.

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