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We are asked to find the limit:

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

### Step-by-Step Solution:

1. **Apply Direct Substitution**:  
   Substituting \(x = 0\) into the expression, we get:
   \[
   \frac{\sin(0)}{\sin(\pi \cdot 0)} = \frac{0}{0}
   \]
   This results in an indeterminate form \( \frac{0}{0} \), so we need to apply **L'Hôpital's Rule**.

2. **L'Hôpital's Rule**:  
   L'Hôpital's Rule states that if a limit results in an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the denominator separately and then find the limit of that expression. Thus, we compute the derivatives of \( \sin(x) \) and \( \sin(\pi x) \).

   - Derivative of \( \sin(x) \) is \( \cos(x) \).
   - Derivative of \( \sin(\pi x) \) is \( \pi \cos(\pi x) \) (using the chain rule).

   Now, apply L'Hôpital's Rule:

   \[
   \lim_{{x \to 0}} \frac{\cos(x)}{\pi \cos(\pi x)}
   \]

3. **Evaluate the New Limit**:  
   Substituting \(x = 0\) into this new expression:

   \[
   \frac{\cos(0)}{\pi \cos(\pi \cdot 0)} = \frac{1}{\pi \cdot 1} = \frac{1}{\pi}
   \]

### Final Answer:
\[
\lim_{{x \to 0}} \frac{\sin(x)}{\sin(\pi x)} = \frac{1}{\pi}
\]

Would you like more details or have any further questions?

Here are 5 related questions to consider:
1. How do you apply L'Hôpital's Rule to more complex indeterminate forms?
2. What are the conditions for using L'Hôpital's Rule?
3. Can you find limits involving different trigonometric functions?
4. How would you solve this limit without L'Hôpital's Rule, using series expansions?
5. How does the limit change if the denominator was \( \sin(kx) \) for any constant \( k \)?

**Tip**: Always check if the limit is indeterminate before applying L'Hôpital's Rule.

Find the limit: lim(x→0) [sin(x)/sin(πx)]

Learn how to solve the limit lim(x→0) [sin(x)/sin(πx)] using L'Hôpital's Rule. This problem explores key trigonometric concepts and the application of derivatives in evaluating limits. Suitable for students in Grades 11-12.

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