MT【236】必要性探路

$\dfrac{lnx}{x+1}+\dfrac{1}{x}>\dfrac{lnx}{x-1}+\dfrac{k}{x}$对于任意$x>0$成立,求$k$的范围.

解答:由题意,对任意$x>0,k<1-\dfrac{2xlnx}{x^2-1}$下面求$\dfrac{2xlnx}{x^2-1}$的最大值.
用洛必达法则
$\lim\limits_{x\longrightarrow 1}\dfrac{2xlnx}{x^2-1}=\lim\limits_{x\longrightarrow 1}\dfrac{2lnx+2}{2x}=1$
下证:$\dfrac{2xlnx}{x^2-1}\le1$构造函数$g(x)=x^2-1-2xlnx$分$x>1,0<x<1$易证.从而$k<0$