用罗必塔法则啊 很简单
对0/0,∞/∞,先对分子分母分别求导,至可以求出极限
lin (x^3-3x+2)/(x-1)^2= lin(3x^2-3)/2(x-1)=lin(6x)/2=+∞
x->+∞ x->+∞ x->+∞
lin 1/x(sinx)+xsin(1/x)]= lin 1/x(sinx)+lin xsin(1/x)
x->+∞ x->+∞ x->+∞
=0+lin((sin(1/x))/(1/x))=0+1=1
x->+∞
因为|1/x(sinx）|=0,故前者极限为0,