Question 14 of 71intermediate🔧 ApplyNumerical1 mark

Evaluate the following limits in Exercises 1 to 22. lim⁡x→πsin⁡(π−x)π(π−x)\lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)}

Correct Answer

1π\frac{1}{\pi}

Exercise: EXERCISE 12.1 | Q: 15 | (Chapter: Page 22)
For More Understanding

Explanation

This question tests the application of the standard limit formula. The student needs to recognize that when x approaches π, the expression (π - x) approaches 0. By substituting y = (π - x), the limit transforms into a form where the standard limit from Theorem 5 can be directly applied.

Solution Steps

  1. Step 1: Let y = π - x. When x → π, we have y → 0.

  2. Step 2: Rewrite the limit: lim⁡x→πsin⁡(π−x)π(π−x)=lim⁡y→0sin⁡yπ⋅y\lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)} = \lim_{y \to 0} \frac{\sin y}{\pi \cdot y}

  3. Step 3: Separate the constant: =1π⋅lim⁡y→0sin⁡yy= \frac{1}{\pi} \cdot \lim_{y \to 0} \frac{\sin y}{y}

  4. Step 4: Apply the standard limit lim⁡y→0sin⁡yy=1\lim_{y \to 0} \frac{\sin y}{y} = 1 (Theorem 5)

  5. Step 5: =1π⋅1=1π= \frac{1}{\pi} \cdot 1 = \frac{1}{\pi}

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