答：
(x→0)lim[sin(x/π)/(x^2-x)] 屬于0-0型,應(yīng)用洛必達(dá)法則：
=(x→0)lim[(1/π)cos(x/π)/(2x-1)]
=(1/π)*(cos0°)/(-1)
=-1/π不對哦，sin(π/x)是有界。再問你lim(1/（x^2-x)）x趨向于0等于請你注明哪個是分母，哪個是分子........分子sin(π/x）分母1/（x^2-x)(x→0)lim[sin(π/x)/(x^2-x)] 因為：-1<=sin(π/x)<=1恒成立而x→0時，1/(x^2-x)趨于無窮大所以：(x→0)lim[sin(π/x)/(x^2-x)] =∞不客氣，祝你學(xué)習(xí)進(jìn)步