题目内容
若
=1,则
=
| lim |
| x→1 |
| f(x-1) |
| x-1 |
| lim |
| x→1 |
| x-1 |
| f(2-2x) |
-
| 1 |
| 2 |
-
.| 1 |
| 2 |
分析:由题意可得
=1,再根据
=
[
(-
)]=-
,运算求得结果.
| lim |
| x→1 |
| 2-2x |
| f(2-2x) |
| lim |
| x→1 |
| x-1 |
| f(2-2x) |
| lim |
| x→1 |
| 2-2x |
| f(2-2x) |
| 1 |
| 2 |
| 1 |
| 2 |
| lim |
| x→1 |
| 2-2x |
| f(2-2x) |
解答:解:∵
=1,∴
=1,
∴
=
[
(-
)]=-
=-
×1=-
,
故答案为-
| lim |
| x→1 |
| f(x-1) |
| x-1 |
| lim |
| x→1 |
| 2-2x |
| f(2-2x) |
∴
| lim |
| x→1 |
| x-1 |
| f(2-2x) |
| lim |
| x→1 |
| 2-2x |
| f(2-2x) |
| 1 |
| 2 |
| 1 |
| 2 |
| lim |
| x→1 |
| 2-2x |
| f(2-2x) |
| 1 |
| 2 |
| 1 |
| 2 |
故答案为-
| 1 |
| 2 |
点评:本题主要考查极限及其运算,式子的变形是解题的关键,属于基础题.
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=1,则
=( )
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