题目内容
| lim |
| x→1 |
| 1 |
| x2-3x+2 |
| 2 |
| x2-4x+3 |
A、-
| ||
B、
| ||
C、-
| ||
D、
|
分析:先利用因式分解和通分,然后再约分,把原式转化为
,由此能求出
(
-
)的值.
| lim |
| x→1 |
| -1 |
| (x-2)(x-3) |
| lim |
| x→1 |
| 1 |
| x2-3x+2 |
| 2 |
| x2-4x+3 |
解答:解:
(
-
)
=
[
-
]
=
=
=-
.
故选A.
| lim |
| x→1 |
| 1 |
| x2-3x+2 |
| 2 |
| x2-4x+3 |
=
| lim |
| x→1 |
| 1 |
| (x-2)(x-1) |
| 2 |
| (x-1)(x-3) |
=
| lim |
| x→1 |
| 1-x |
| (x-1)(x-2)(x-3) |
=
| lim |
| x→1 |
| -1 |
| (x-2)(x-3) |
=-
| 1 |
| 2 |
故选A.
点评:本题考查
型函数的极限问题,解题时要注意消除零因子.
| 0 |
| 0 |
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