题目内容
| lim |
| n→∞ |
| 1 |
| n2+1 |
| 3 |
| n2+1 |
| 2n-1 |
| n2+1 |
分析：根据同分母分式加法的性质和等差数列的求各公式，原式可以等价转化为
=
，由此能够求出其极限值．
| lim |
| n→∞ |
| ||
| n2+1 |
| lim |
| n→∞ |
| n2 |
| n2+1 |
解答：解：
(
+
+…+
)=
=
=1，
故答案为1．
| lim |
| n→∞ |
| 1 |
| n2+1 |
| 3 |
| n2+1 |
| 2n-1 |
| n2+1 |
| lim |
| n→∞ |
| ||
| n2+1 |
| lim |
| n→∞ |
| n2 |
| n2+1 |
故答案为1．
点评：本题考查
型极限的求法，解题的关键是正确选取公式．
| ∞ |
| ∞ |
练习册系列答案
相关题目