Discussion

The limit

\[ \begin{aligned} p &= \lim_{x \to \pi} \\ &\left(\frac{x^{2} + \alpha x + 2\pi^{2}}{x - \pi + 2\sin x}\right) \end{aligned} \]

has a finite value for a real \(\alpha\). The value of \(\alpha\) and the corresponding limit \(p\) are

Answer: A

The values are \(\alpha = -3\pi\) and \(p = \pi\).

At \(x = \pi\) the denominator is \(\pi - \pi + 2\sin\pi = 0\). For the limit to be finite, the numerator must vanish there too:

\[ \begin{aligned} \pi^{2} + \alpha\pi + 2\pi^{2} &= 0 \\ \alpha &= -3\pi \end{aligned} \]

The limit is then of the form 0/0, and L'Hôpital's rule gives

\[ \begin{aligned} p &= \lim_{x \to \pi}\frac{2x + \alpha}{1 + 2\cos x} \\ &= \frac{2\pi - 3\pi}{1 - 2} = \pi \end{aligned} \]

Any other \(\alpha\) leaves a non-zero numerator over a zero denominator, so the limit would be infinite; that rules out the other three options.

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