题目内容
（1）计算| lim |
| n→∞ |
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| 4n2 |
（2）若
| lim |
| n→∞ |
| an2-2n+1 |
| bn+2 |
| a |
| b |
分析：（1）(1-
)(1-
)(1-
)…(1-
)=
•
=
，由此能求出其结果．
（2）2n+
=
，且
(2n+
)=1，由此能求出
=-2．
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| 4n2 |
| 1 |
| 2 |
| 2n+1 |
| 2n |
| 2n+1 |
| 4n |
（2）2n+
| an2-2n+1 |
| bn+2 |
| (2b+a)n2+2n+1 |
| bn+2 |
| lim |
| n→∞ |
| an2-2n+1 |
| bn+2 |
| a |
| b |
解答：解：（1）(1-
)(1-
)(1-
)…(1-
)=
•
=
，
所以
(1-
)(1-
)(1-
)…(1-
)=
=
．
（2）2n+
=
，
且
(2n+
)=1，
所以
，
即
=-2．
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| 4n2 |
| 1 |
| 2 |
| 2n+1 |
| 2n |
| 2n+1 |
| 4n |
所以
| lim |
| n→∞ |
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| 4n2 |
| lim |
| n→∞ |
| 2n+1 |
| 4n |
| 1 |
| 2 |
（2）2n+
| an2-2n+1 |
| bn+2 |
| (2b+a)n2+2n+1 |
| bn+2 |
且
| lim |
| n→∞ |
| an2-2n+1 |
| bn+2 |
所以
|
即
| a |
| b |
点评：本题考查极限的性质和运算，解题时要认真审题，仔细思考，注意合理地进行等价转化．
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