思路如图：
将x=-1代入上式即可得极限，-1
(x^2-x+1-3)/(x^3+1)=(X^2-x-2)/(x^3+1)=(x-2)/(x^2-x+1);
极限-1。
1/(x+1)－3/(x^3+1)＝[(x^2-x+1)-2]/(x^3+1)＝(x-2)/(x^2-x+1)，极限是(1-2)/(1-1+1)＝-1
求[1\/(x+1)]-[3\/(x^3+1)]当x趋近于-1时的极限
思路如图：
[1\/(x+1)-3\/(x^3+1)]当x无限接近-1时的极限?
首先x^3+1=(x+1)(x^2+1-x) 则通分变原式为x^2+1-x-3\/(x+1)(x^2+1-x)=（x-2)(x+1)\/(x+1)(x^2+1-x)=x-2\/x^2+1-x 所以当x趋近-1时 其极限为-3\/3=-1 采纳哦
求lim [(1\/x+1)-(3\/x^3+1)]= x→-1
简单分析一下，详情如图所示
...急~~ lim(1\/x+1-3\/X^3+1) x趋近-1 求函数极限
通分,分解因式,约去X+1 lim(x^2-x+1-3)\/(X^3+1)=lim(x^2-x-2)\/\/X^3+1)=lim(x-2)\/(x^2-x+1)=-1
...急~~ lim(1\/x+1-3\/X^3+1) x趋近-1 求函数极限
通分,分解因式,约去X+1 lim(x^2-x+1-3)\/(X^3+1)=lim(x^2-x-2)\/\/X^3+1)=lim(x-2)\/(x^2-x+1)=-1
{1\/(1-x)}-{3\/(1-x的三次方)}当x趋近于一的时候的极限是多少
1\/(1-x)-3\/(1-x^3)=(1+x+x^2-3)\/(1-x^3)=(x^2+x-2)\/(1-x^3)=(x+2)(x-1)\/(1-x)(1+x+x^2)=-(x+2)\/(1+x+x^2)所以x趋近于1时，极限=-(1+2)\/(1+1+1)=-1
lim(x→-1) [3\/(x^3+1)-1\/(x+1)]
回答：x3+1=(x+1)(x2-x+1)
当x趋近于-1时3\/(x^3+1)-1\/(x+1)的极限
😁
设lim(x→-1)[1\/(ax+1)-3\/(x^3+1)]=b
如图所示
求x趋于1时 [ 3\/(1-x^3) ]-1\/(1-x)的极限
lim(x→-1)[1\/(x+1)-3\/(x^3+1)]=lim(x→-1)[(x^2-x+1-3)\/(x^3+1)]=lim(x→-1)[(x^2-x-2)\/(x^3+1)]=lim(x→-1)[(x+1)(x-2)\/(x+1)(x^2-x+1)]=lim(x→-1)[(x-2)\/(x^2-x+1)]=-3\/3 =-1 ...
