| 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) |