$dfrac{lnx}{x+1}+dfrac{1}{x}>dfrac{lnx}{x-1}+dfrac{k}{x}$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
$limlimits_{xlongrightarrow 1}dfrac{2xlnx}{x^2-1}=limlimits_{xlongrightarrow 1}dfrac{2lnx+2}{2x}=1$
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$k<0$