{Najdi limitu
$$
   \lim_{x\to\pi/6}\frac{2\sin^2 x+\sin x - 1}{2\sin^2 x-3\sin x + 1}.
$$}{
\begin{eqnarray*}
   \lim_{x\to\pi/6}\frac{2\sin^2 x+\sin x - 1}{2\sin^2 x-3\sin x + 1} & = &
   \lim_{x\to\pi/6}\frac{(2\sin x-1)(\sin x+1)}{(2\sin x-1)(\sin x-1)}\\
   & = & \lim_{x\to\pi/6}\frac{\sin x+1}{\sin x-1}\\
   & = & \frac{3/2}{-1/2}\;=\;-3.
\end{eqnarray*}
}
