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 |
所以有水平漸近線y=
π |
4 |
垂直漸近線:
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) |
所以有水平漸近線x=0,x=-1,x=2
不存在斜漸近線
故應(yīng)選: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) |