Math Problem Statement

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

Solution

We are asked to find the limit:

limx0sin(x)sin(πx)\lim_{{x \to 0}} \frac{\sin(x)}{\sin(\pi x)}

Step-by-Step Solution:

  1. Apply Direct Substitution:
    Substituting x=0x = 0 into the expression, we get: sin(0)sin(π0)=00\frac{\sin(0)}{\sin(\pi \cdot 0)} = \frac{0}{0} This results in an indeterminate form 00\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 00\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)\sin(x) and sin(πx)\sin(\pi x).

    • Derivative of sin(x)\sin(x) is cos(x)\cos(x).
    • Derivative of sin(πx)\sin(\pi x) is πcos(πx)\pi \cos(\pi x) (using the chain rule).

    Now, apply L'Hôpital's Rule:

    limx0cos(x)πcos(πx)\lim_{{x \to 0}} \frac{\cos(x)}{\pi \cos(\pi x)}

  3. Evaluate the New Limit:
    Substituting x=0x = 0 into this new expression:

    cos(0)πcos(π0)=1π1=1π\frac{\cos(0)}{\pi \cos(\pi \cdot 0)} = \frac{1}{\pi \cdot 1} = \frac{1}{\pi}

Final Answer:

limx0sin(x)sin(πx)=1π\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)\sin(kx) for any constant kk?

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

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
Indeterminate Forms

Formulas

L'Hôpital's Rule: If a limit results in 0/0 or ∞/∞, take the derivative of the numerator and denominator separately
Derivative of sin(x) = cos(x)
Derivative of sin(πx) = πcos(πx)

Theorems

L'Hôpital's Rule

Suitable Grade Level

Grades 11-12