Question

I need help on this problem please ?

Answer 1

The answer to this answer is B

Answer 2

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.

$\lim_{x \to \pi} (\frac{-sinx + 2cos(2x)}{2x})$
Now, substituting pi into the equation renders $\frac{0 + 2}{2\pi}$

So, our limit becomes $\frac{1}{\pi}$ (ie B)