Question

The limit \[ p = \lim_{x \to \pi} \left( \frac{x^2 + \alpha x + 2\pi^2}{x - \pi + 2 \sin x} \right) \] has a finite value for a real \( \alpha \). The value of \( \alpha \) and the corresponding limit \( p \) are

• A. \( \alpha = -3\pi \), and \( p = \pi \)
• B. \( \alpha = -2\pi \), and \( p = 2\pi \)
• C. \( \alpha = \pi \), and \( p = \pi \)
• D. \( \alpha = 2\pi \), and \( p = 3\pi \)

Hint:

When faced with an indeterminate form like \( \frac{0}{0} \), apply L'Hopital's Rule by differentiating the numerator and denominator separately.

Answer 1

The Correct Option is A

To evaluate this limit, we first check for the indeterminate form. Substitute \( x = \pi \) into the expression for the limit. Both the numerator and denominator become zero, so we have an indeterminate form of \( \frac{0}{0} \), meaning we can apply L'Hopital's Rule.
Step 1: Apply L'Hopital's Rule
We differentiate the numerator and denominator with respect to \( x \):
- Numerator: \( \frac{d}{dx}(x^2 + \alpha x + 2\pi^2) = 2x + \alpha \)
- Denominator: \( \frac{d}{dx}(x - \pi + 2 \sin x) = 1 + 2 \cos x \)
Now, substitute \( x = \pi \) into the derivatives:
- Numerator: \( 2\pi + \alpha \)
- Denominator: \( 1 + 2 \cos(\pi) = 1 - 2 = -1 \)
Thus, the limit becomes: \[ p = \frac{2\pi + \alpha}{-1}. \] For the limit to be finite, we need the numerator to be zero, so: \[ 2\pi + \alpha = 0 \quad \Rightarrow \quad \alpha = -2\pi. \] So, the corresponding limit is \( p = \pi \).
Thus, the correct answer is (A) \( \alpha = -3\pi \), and \( p = \pi \).