Forgot password
 Register account
View 313|Reply 7

[函数] 如何不用洛必达求lnx*[(x+1)/(x-1)]最小值

[Copy link]

41

Threads

60

Posts

0

Reputation

Show all posts

baxiannv posted 2022-4-30 19:53 |Read mode
Last edited by baxiannv 2022-4-30 20:12定义域(1,+∞)

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2022-4-30 20:03
请检查题目

41

Threads

60

Posts

0

Reputation

Show all posts

original poster baxiannv posted 2022-4-30 20:12
[⊙∀⊙! 打错了x-1b]回复 2# kuing [/b]

3200

Threads

7827

Posts

52

Reputation

Show all posts

hbghlyj posted 2022-4-30 20:28
回复 2# kuing

他应该是想问$$\lim_{x→1}\frac{(x+1) \log (x)}{x-1}=2$$
$$⇔\underset{x\to 0}{\text{lim}}\frac{\left(e^x+1\right) x}{e^x-1}=2$$
$$⇔\underset{x\to 0}{\text{lim}}\frac{x}{\tanh \left(\frac{x}{2}\right)}=2$$
$$⇔\underset{x\to 0}{\text{lim}}\frac{x}{\tanh x}=1$$

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2022-4-30 21:34
那到底是求极限还是求最小值……

或者说,原题是不是那种带参数的函数恒成立啥的然后求参数范围?

3200

Threads

7827

Posts

52

Reputation

Show all posts

hbghlyj posted 2022-4-30 22:55
回复 5# kuing

求导容易得到lnx*[(x+1)/(x-1)]在(1,+∞)上单调增,他求最小值时需要计算x→1的极限

764

Threads

4672

Posts

27

Reputation

Show all posts

isee posted 2022-4-30 22:57
回复 6# hbghlyj


这就是没有最小值啊,洛必达法则太无辜了


==今天帖子过百了,你(们)太狠了

3200

Threads

7827

Posts

52

Reputation

Show all posts

hbghlyj posted 2022-4-30 22:58
回复 7# isee
下确界是2

Quick Reply

Advanced Mode
B Color Image Link Quote Code Smilies
You have to log in before you can reply Login | Register account

$\LaTeX$ formula tutorial

Mobile version

2025-7-15 14:24 GMT+8

Powered by Discuz!

Processed in 0.014338 seconds, 23 queries