show that x >lnx for every x>0
please help~!!
自然對數 證明題
2008-01-07 2:00 pm
回答 (2)
2008-01-07 9:34 pm
✔ 最佳答案
let f(x) = x -lnxf(x)= ln(e^x) - ln(x)
=ln(e^x/x)
since (e^x) = lim(1+x/n)^n, n tends to + infinity
=(1+x+x^2/2!+x^3/3!+......)
therefore
f(x)=ln(e^x/x)=ln( 1+x+x^2/2!+......)/x
＞ln[(1+x)/x] ＞ln(1)=0
hence x-lnx＞０
x＞lnx
(這證法是從e的定義出發）
2008-01-07 3:19 pm
let f(x)=x-lnx
then f'(x)=1-1/x>0 for all x>1
and f'(x)<0 for all 0<1
so f(x) is min. when x=1
so f(x)>f(0)
x- lnx>1-ln1
x>1+lnx>lnx
then f'(x)=1-1/x>0 for all x>1
and f'(x)<0 for all 0<1
so f(x) is min. when x=1
so f(x)>f(0)
x- lnx>1-ln1
x>1+lnx>lnx
收錄日期: 2021-04-11 16:21:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080107000051KK00456