题目内容
(2000•上海)计算:
(
)n=
| lim |
| n→∞ |
| n |
| n+2 |
e-2
e-2
.分析:将极限转化为
(1+
)x型,利用极限公式
(1+
)x=e,进行求值.
| lim |
| x→∞ |
| 1 |
| x |
| lim |
| x→∞ |
| 1 |
| x |
解答:解:因为(
)n=
=
,所以
(
)n=
=
=e-2.
故答案为:e-2.
| n |
| n+2 |
| 1 | ||
(1+
|
| 1 | ||||||
[(1+
|
| lim |
| n→∞ |
| n |
| n+2 |
| lim |
| n→∞ |
| 1 | ||||||
[(1+
|
| 1 |
| e2 |
故答案为:e-2.
点评:本题考查的数列极限的求法,将所求极限转化为几种常见极限的类型是解决本题的关键.要求熟练掌握各种求极限的极限公式.
练习册系列答案
相关题目