lim |
x→∞ |
1 |
x2 |
x2+x−1 |
(x+1)(x−2) |
lim |
x→∞ |
1 |
x2 |
lim |
x→∞ |
x2+x−1 |
(x+1)(x−2) |
π |
4 |
所以有水平渐近线y=
π |
4 |
垂直渐近线:
lim |
x→0 |
1 |
x2 |
x2+x−1 |
(x+1)(x−2) |
lim |
x→1− |
1 |
x2 |
x2+x−1 |
(x+1)(x−2) |
lim |
x→2+ |
1 |
x2 |
x2+x−1 |
(x+1)(x−2) |
所以有水平渐近线x=0,x=-1,x=2
不存在斜渐近线
故应选:D.
1 |
x2 |
x2+x−1 |
(x+1)(x−2) |
lim |
x→∞ |
1 |
x2 |
x2+x−1 |
(x+1)(x−2) |
lim |
x→∞ |
1 |
x2 |
lim |
x→∞ |
x2+x−1 |
(x+1)(x−2) |
π |
4 |
π |
4 |
lim |
x→0 |
1 |
x2 |
x2+x−1 |
(x+1)(x−2) |
lim |
x→1− |
1 |
x2 |
x2+x−1 |
(x+1)(x−2) |
lim |
x→2+ |
1 |
x2 |
x2+x−1 |
(x+1)(x−2) |