题目内容
计算:| lim |
| n→∞ |
| ||
| n3+1 |
分析:
=
,由此能够导出:
.
| lim |
| n→∞ |
| ||
| n3+1 |
| lim |
| n→∞ |
| n3-3n2+2n |
| (n3+1) •3! |
| lim |
| n→∞ |
| ||
| n3+1 |
解答:解:
=
=
=
=
.
答案:
.
| lim |
| n→∞ |
| ||
| n3+1 |
| lim |
| n→∞ |
| n(n-1)(n-2) |
| (n3+1)•3! |
| lim |
| n→∞ |
| n3-3n2+2n |
| (n3+1)•3! |
| lim |
| n→∞ |
1-
| ||||
(1+
|
| 1 |
| 6 |
答案:
| 1 |
| 6 |
点评:本题考查组合和极限的基本性质,解题时要认真审题,仔细解除,注意公式的灵活运用.
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