请
这类型题的思想都是:通分,再分解因式,约去分母中可能为0的因式即可。本题这样处理的:因为:x^3-1=(x-1)(x^2+x+1) ,所以通分化简,有:1/(x-1)-3/(x^3-1)=(x^2+x+1-3)/(x^3-1)=(x+2)(x-1)/(x-1)(x^2+x+1)=(x+2)/(x^2+x+1) ;则所求极限等于: (1+2)/(1^2+1+1)=1
lim(x->1) [ 1/(x-1) - 3/(x^3-1) ]
=lim(x->1) [(x^2+x+1) -3 ]/[(x-1)(x^2+x+1)]
=lim(x->1) (x^2+x-2)/[(x-1)(x^2+x+1)]
=lim(x->1) (x-1)(x+2)/[(x-1)(x^2+x+1)]
=lim(x->1) (x+2)/(x^2+x+1)
=(1+2)/(1+1+1)
=1