Question

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In the attached image is a AP calculus problem.

Answer 1

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

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

Answer 2

`B. \ (1)/(\pi)`

Explanation: