Use L'Hopital's Rule.
Since substituting pi into the equation renders 0/0, we need to take the derivative of both, the numerator and denominator and then using the limits again.
[tex]\lim_{x \to \pi} (\frac{-sinx + 2cos(2x)}{2x})[/tex]
Now, substituting pi into the equation renders [tex]\frac{0 + 2}{2\pi}[/tex]
So, our limit becomes [tex]\frac{1}{\pi}[/tex] (ie B)
