lim(x→0+)[In1/(x+1)]/x=?请问这个怎么换算到lim(x→0+)[In1/(x+1)]/x= -lim(x→0+)[In(x+1)]/x= -1怎么等于-1的我也没看懂
lim(x→0+)[In1/(x+1)]/x=?
请问这个怎么换算到
lim(x→0+)[In1/(x+1)]/x
= -lim(x→0+)[In(x+1)]/x
= -1
怎么等于-1的我也没看懂
lim(x→0+)[In1/(x+1)]/x=?
答案:2 悬赏:80 手机版
解决时间 2021-12-28 15:49
- 提问者网友：别再叽里呱啦
- 2021-12-27 15:35
最佳答案
- 五星知识达人网友：狂恋
- 2022-01-06 09:18
In1/(x+1)
=ln(x+1)^(-1)
=-ln(x+1)
所以lim(x→0+)[In1/(x+1)]/x
= -lim(x→0+)[In(x+1)]/x
然后用洛必达法则
[ln(x+1)]'=1/(x+1)
所以lim(x→0+)[In1/(x+1)]/x
= -lim(x→0+)[In(x+1)]/x
= -lim(x→0+)1/(x+1)
=-1
=ln(x+1)^(-1)
=-ln(x+1)
所以lim(x→0+)[In1/(x+1)]/x
= -lim(x→0+)[In(x+1)]/x
然后用洛必达法则
[ln(x+1)]'=1/(x+1)
所以lim(x→0+)[In1/(x+1)]/x
= -lim(x→0+)[In(x+1)]/x
= -lim(x→0+)1/(x+1)
=-1
全部回答
- 1楼网友：忘川信使
- 2022-01-06 09:45
-1
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