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To evaluate the limit using L'Hôpital's Rule, let's proceed with the given expression:

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

### Step 1: Verify the Indeterminate Form
As \( x \to 0 \):
- The numerator \(\sin 5x \to 0\),
- The denominator \(x \to 0\).

This creates the indeterminate form \( \frac{0}{0} \), which allows us to apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, if we have an indeterminate form \(\frac{0}{0}\), we can differentiate the numerator and the denominator separately:

1. Differentiate the numerator: \(\frac{d}{dx}(\sin 5x) = 5 \cos 5x\).
2. Differentiate the denominator: \(\frac{d}{dx}(x) = 1\).

The limit then becomes:

\[
\lim_{{x \to 0}} \frac{5 \cos 5x}{1}
\]

### Step 3: Substitute \( x = 0 \)
Now, substitute \( x = 0 \):

\[
5 \cos 5(0) = 5 \cos 0 = 5 \times 1 = 5
\]

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

Would you like further clarification or details on any of the steps? 

### Related Questions:
1. How would the limit change if we evaluated \(\lim_{{x \to 0}} \frac{\sin kx}{x}\) for any constant \(k\)?
2. Can we apply L'Hôpital's Rule to solve \(\lim_{{x \to 0}} \frac{\sin x}{x^2}\)?
3. What is the result of \(\lim_{{x \to 0}} \frac{\tan 5x}{x}\)?
4. Why is L'Hôpital's Rule applicable only in certain indeterminate forms?
5. How does the result differ if we calculate \(\lim_{{x \to 0}} \frac{\sin 5x}{2x}\)?

### Tip:
When using L'Hôpital's Rule, always verify that the limit initially presents an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) before proceeding.

Use L'Hôpital's Rule to evaluate the limit: lim (x→0) (sin 5x) / x

Learn how to evaluate the limit lim (x→0) (sin 5x) / x using L'Hôpital's Rule. This solution provides a step-by-step explanation suitable for calculus students, including differentiation and substitution to find the final result.

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