{2ln(x)*(1/x)}/{2x} = {ln(x)}/{x}*(1/x) --->(x-->infty) 0*0=0

In second problem:

n*sin(n)=sin(1/n)/{1/n} --->(n-->infty) sin(0)/(0)=0/0 {sinx is continuous function}

We ca use L'Hopital's rule! (WHY?)

{sin(1/n)}'/{1/n}'={cos(1/n)*(-1/n^2)}/{(-1/n^2}=cos(1/n) --->(n-->infty) cos(0)=1 {cosx is continuous function}