Question

当x→-1的时候1/(x+1)-3/(x^3+1)的极限是多少

Answer 1

设f(x)=1/(x＋1)－3/(x³＋1)=(x²－x－2)/(x³＋1)
则对f(x)分子分母分别求导，得：(2x－1)/(3x²)，
则：原来的极限=分子分母分别求导后的极限==>>>以x=－1代入== －1

Answer 2

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

追问

lim(x→-1)[(x^2-x+1-3)/(x^3+1)]
如何得到

Answer 3

limx->-1 (x^-x+1)/(x^3+1)-3/(x^3+1)
=limx->-1 (x^2-x-2)/(x^3+1)
=limx->-1 (x-2)(x+1)/(x^3+1)
=limx->-1 (x-2)/(x^2-x+1)
=-3/3=-1
觉得好请采纳 谢谢

Answer 4

1/(x+1)-3/(x^3+1)=(x^2-x-2)/(x+1)(x^2-x+1)=(x-2)(x+1)/(x+1)(x^2-x+1)=（x-2)/(x^2-x+1)
所以上式在x趋于-1时的极限为-3/4

Answer 5

In[1]:=
Limit[1/(x+1)-3/(x^3+1),x->1]
Out[1]=
-1

用Mathematica 算得。