證明lim(x->0) ( a^x-1)/x

2014-06-20 3:47 am
現在證明了
lim(x->0) log<a>(1+x) / x =log<a>e

那麼如何證明lim(x->0) ( a^x-1)/x=ln e??

請不要用洛必達法則來做。

多謝!!

回答 (2)

2014-06-20 3:58 am
✔ 最佳答案
題目不正確吧?

應該是 ln a ?

2014-06-19 19:58:05 補充:
Please read:


圖片參考:https://s.yimg.com/rk/HA00430218/o/873939996.png


2014-06-19 23:02:42 補充:
HK~

不要以為 L'Hopital rule 是無敵。

請留意兩個情況:

lim(x→0) (e^x - 1)/x = 1 如果你用洛必達法則你必定犯了循環論證的錯誤。
請研究研究~

另外,洛必達法則也未必一定是最佳方法:
https://tw.knowledge.yahoo.com/question/question?qid=1614061106437

2014-06-19 23:15:10 補充:
哦~
原來係咁~

我以為你的意思係想去用 L'Hopital rule 去 prove 添~

☆ヾ(◕‿◕)ノ

2014-06-19 23:32:11 補充:
聽講有可能有其他解法:

https://tw.knowledge.yahoo.com/question/question?qid=1014061200420

2014-06-19 23:34:25 補充:
BTW,你而家被人投緊票果題有人玩你,我唔想你的心機白費,你快d去處理一下。

(◕‿◕✿)

2014-06-20 00:04:48 補充:
That is a wrong method!!!
This involves circular argument.

This is because (e^x - 1)/x → 1 as x → 0 is needed to prove that (e^x)' = e^x.

However, now you use (e^x)' = e^x to prove (e^x - 1)/x → 1 as x → 0.

2014-06-20 00:07:22 補充:
請好好學習此帖:

https://tw.knowledge.yahoo.com/question/question?qid=1513071005269

2014-06-20 00:54:23 補充:
我知,但我講緊 012 的帖。
2014-06-20 6:59 am
洛必達法則 -.-

2014-06-19 23:12:50 補充:
其實我無諗d咩...

只係覺得個中文名好怪 0.0

2014-06-19 23:31:03 補充:
話說頭先哥個網嘅哥1題,

如果唔用power series去計嘅話,

都幾麻煩 0.0

2014-06-19 23:39:50 補充:
....

你唔講我都唔記得tim..

thx,過去睇下先

2014-06-19 23:58:04 補充:
謝謝你~

btw,我係yahoo! search哥到搵到以下嘅1個網 ...

https://answers.yahoo.com/question/index;_ylt=A2oKmJeAAqNTlS0ALeizygt.;_ylu=X3oDMTBzazlzZnBnBHNlYwNzcgRwb3MDMTAEY29sbwNzZzMEdnRpZAM-?qid=20080930134940AAVSfbZ

2014-06-20 00:09:23 補充:
好彩我係用power series去做 -.-''

2014-06-20 00:36:47 補充:
https://tw.knowledge.yahoo.com/question/question?qid=1014061200420

呢題用L'Hopital rule要d 3次....

2014-06-20 13:53:06 補充:
i know ...

哥個帖用左4個比較好嘅方法去prove lim[x->0] (e^x-1)/x

亦有說明用L'Hopital去證係錯嘅

也證明了L'Hopital不是萬能


收錄日期: 2021-04-11 20:42:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140619000051KK00124

檢視 Wayback Machine 備份