Вычислить предел!!Помогите

Решите задачу:

$\displaystyle\lim _{x\to \:\pi }\left(\frac{\sin ^2\left(x\right)}{1+\cos \left(x\right)}\right) =\lim _{x\to \:\pi }\left(\frac{\sin \left(2x\right)}{-\sin \left(x\right)}\right)=\lim _{x\to \:\pi }\left(\frac{-\sin \left(2x\right)}{\sin \left(x\right)}\right)= \\ \\ =\lim _{x\to \:\pi }\left(\frac{-\cos \left(2x\right)\cdot \:2}{\cos \left(x\right)}\right)=\frac{-\cos \left(2\pi \right)\cdot \:2}{\cos \left(\pi \right)}=\frac{-\left(-1+2\cos ^2\left(\pi \right)\right)\cdot \:2}{\cos \left(\pi \right)}=$
$\displaystyle\frac{2\left(-1+2\cos ^2\left(\pi \right)\right)}{\cos \left(\pi \right)} =-\frac{2\left(-1+2\left(-1\right)^2\right)}{-1}=-\left(-\frac{2}{1}\right)=\frac{2}{1}=2$