题目内容
计算:
[n(
-
)]= .
| lim |
| n→∞ |
| 2 |
| n |
| 1 |
| n+1 |
分析:由极限的性质,把
[n(
-
)]=
(2-
)等价转化为
(2-
),由此能够求出结果.
| lim |
| n→∞ |
| 2 |
| n |
| 1 |
| n+1 |
| lim |
| n→∞ |
| n |
| n+1 |
| lim |
| n→∞ |
| 1 | ||
1+
|
解答:解:
[n(
-
)]=
(2-
)=
(2-
)=2-1=1.
故答案为:1.
| lim |
| n→∞ |
| 2 |
| n |
| 1 |
| n+1 |
| lim |
| n→∞ |
| n |
| n+1 |
| lim |
| n→∞ |
| 1 | ||
1+
|
故答案为:1.
点评:本题考查极限的性质和应用,是基础题.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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