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Решите задачу:

$\lim\limits _{x\to \frac{\pi}{6}} \frac{2sin^2x+sinx-1}{2sin^2x-3sinx+1} =\\\\=[\, 2t^2+t-1=2(t+1)(t-\frac{1}{2})\; ,\; \; 2t^2-3t+1=2(t-1)(t-\frac{1}{2})\, ]=\\\\=\lim\limits _{x\to \frac{\pi}{6}} \frac{2(sinx+1)(sinx-\frac{1}{2})}{2(sinx-1)(sinx-\frac{1}{2})} =\lim\limits _{\x\to \frac{\pi}{6}} \frac{sin\frac{\pi}{6}+1}{sin\frac{\pi}{6}-1} = \frac{\frac{1}{2}+1}{\frac{1}{2}-1} = \frac{3/2}{-1/2} =-3$