题目内容
已知则
=2,则a+b=
| lim |
| x→2 |
| x2+ax+b |
| x2-x-2 |
-6
-6
.分析:由
=2,知4+2a+b=0,所以
=2等价转化为
=
,由此能求出a+b.
| lim |
| x→2 |
| x2+ax+b |
| x2-x-2 |
| lim |
| x→2 |
| x2+ax+b |
| x2-x-2 |
| lim |
| x→2 |
| x2+ax-2a-4 |
| (x+1)(x-2) |
| lim |
| x→2 |
| (x-2)(x+2+a) |
| (x+1)(x-2) |
解答:解:∵
=2,
∴4+2a+b=0,
∴
=2能够转化为
=
=
=2,
∴
=2,
∴a=2,b=-8,
∴a+b=-6.
故答案为:-6.
| lim |
| x→2 |
| x2+ax+b |
| x2-x-2 |
∴4+2a+b=0,
∴
| lim |
| x→2 |
| x2+ax+b |
| x2-x-2 |
| lim |
| x→2 |
| x2+ax-2a-4 |
| (x+1)(x-2) |
=
| lim |
| x→2 |
| (x-2)(x+2+a) |
| (x+1)(x-2) |
=
| lim |
| x→2 |
| x+2+a |
| x+1 |
∴
| 4+a |
| 3 |
∴a=2,b=-8,
∴a+b=-6.
故答案为:-6.
点评:本题考查极限的性质和应用,解题时要认真审题,仔细解答,注意合理地进行等价转化.
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