Since 0<x<y, y>x, y/x>1. ln(y)-ln(x)=ln(y/x)>0; y-x=x(y/x-1).
(ln(y)-ln(x))/(y-x)=ln(y/x)/(y-x).
Let y/x=1+p, so ln(y/x)=ln(1+p)=p-p2/2+p3/3-...;
Consider x→y, then p is small, so ln(y/x)=ln(1+p)<p, that is, <y/x-1, <(y-x)/x.
So (ln(y)-ln(x))/(y-x)→[(y-x)/x]/(y-x)<1/x.
Consider x=y/n, where n>1, then y/x=n,
so (ln(y)-ln(x))/(y-x)=ln(n)/(nx-x)=ln(n)/[(y/n)(n-1))]=nln(n)/(y(n-1)).
When n is very large n/(n-1)→1 and x→0 so (ln(y)-ln(x))/(y-x)→ln(n)/y>1/y because ln(n)>>1 when is large. Therefore,
1/y<(ln(y)-ln(x))/(y-x)<1/x.