题目内容
已知
=2,
=3,则
=( )
| lim |
| n→∞ |
| an2+cn |
| bn2+c |
| lim |
| n→∞ |
| bn+c |
| cn+a |
| lim |
| n→∞ |
| an2+bn+c |
| cn2+an+b |
A、
| ||
B、
| ||
C、
| ||
| D、6 |
分析:由题意可得
=2,
=3,从而得到
=6,利用数列极限的运算法则把要求的式子化为
=
,由此求得结果.
| a |
| b |
| b |
| c |
| a |
| c |
| lim |
| n→∞ |
a +
| ||||
c+
|
| a |
| c |
解答:解:∵
=2,
=3,∴
=2,
=3,∴
=2×3=6.
∴
=
=
=
=6,
故选D.
| lim |
| n→∞ |
| an2+cn |
| bn2+c |
| lim |
| n→∞ |
| bn+c |
| cn+a |
| a |
| b |
| b |
| c |
| a |
| c |
∴
| lim |
| n→∞ |
| an2+bn+c |
| cn2+an+b |
| lim |
| n→∞ |
a +
| ||||
c+
|
| a+0+0 |
| c+0+0 |
| a |
| c |
故选D.
点评:本题考查数列极限的运算法则,由条件求得
=6,是解题的关键.
| a |
| c |
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