Tính \(\lim\limits_{x\rightarrow\dfrac{\pi}{6}}\dfrac{2\sin^2x+\sin x-1}{2\sin^2x-3\sin x+1}\).
\(3\).\(-3\).\(\dfrac{1}{3}\).\(-\dfrac{1}{3}\).Hướng dẫn giải:Có \(\lim\limits_{x\rightarrow\dfrac{\pi}{6}}\dfrac{2\sin^2x+\sin x-1}{2\sin^2x-3\sin x+1}=\lim\limits_{x\rightarrow\dfrac{\pi}{6}}\dfrac{2\left(\sin x+1\right)\left(\sin x-\dfrac{1}{2}\right)}{2\left(\sin x-1\right)\left(\sin x-\dfrac{1}{2}\right)}=\dfrac{\sin x+1}{\sin x-1}\) nên giới hạn cần tính bằng \(\lim\limits_{x\rightarrow\dfrac{\pi}{6}}\dfrac{\sin x+1}{\sin x-1}=-3\)
