答：
(x→0)lim[sin(x/π)/(x^2-x)] 属于0-0型,应用洛必达法则：
=(x→0)lim[(1/π)cos(x/π)/(2x-1)]
=(1/π)*(cos0°)/(-1)
=-1/π不对哦，sin(π/x)是有界。再问你lim(1/（x^2-x)）x趋向于0等于请你注明哪个是分母，哪个是分子........分子sin(π/x）分母1/（x^2-x)(x→0)lim[sin(π/x)/(x^2-x)] 因为：-1<=sin(π/x)<=1恒成立而x→0时，1/(x^2-x)趋于无穷大所以：(x→0)lim[sin(π/x)/(x^2-x)] =∞不客气，祝你学习进步