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In the attached image is a AP calculus problem.

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Answer:

[tex]B. \ \frac{1}{\pi}[/tex]

Step-by-step explanation:

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You also have l'Hopitâl's rule at your disposal.

[tex]\displaystyle \lim_{x\to\pi} \frac{\cos(x) + \sin(2x) + 1}{x^2 - \pi^2} = \lim_{x\to\pi} \frac{-\sin(x) + 2\cos(2x)}{2x} = \frac{0+2\cos(2\pi)}{2\pi} = \boxed{\frac1\pi}[/tex]

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