Use L'Hôpital's rule.
If #lim_(x->c)f(x)/g(x)# is one of the indeterminate forms #0/0# or #oo/oo#, then #lim_(x->c)f(x)/g(x)=lim_(x->c)(f'(x))/(g'(x))#, where #'# means the derivative with respect to #x#.
#lim_(x->oo)ln^2(x)/x# is of the form #oo/oo#. Thus, #lim_(x->oo)(d/dx(ln^2(x)))/(d/dx(x))=lim_(x->oo)(2ln(x))/x#.
This is still of the form #oo/oo#. Apply L'Hôpital's rule again: #lim_(x->oo)(2ln(x))/x=lim_(x->oo)(d/dx(2ln(x)))/(d/dx(x))=lim_(x->oo)2/x=0#.
This graph demonstrates that #ln^2(x)/x# does approach #0# as #x# approaches #oo# (very slowly).
graph{ln(x)^2/x [-8.89, 8.89, -4.446, 4.446]}
