The limit:
#lim_(x->oo) (lnx)^2/sqrt(x)#
is in the indeterminate form: #oo/oo# so we can use l'Hospital's rule:
#lim_(x->oo) (lnx)^2/sqrt(x) = lim_(x->oo) (d/dx(lnx)^2)/(d/dxsqrt(x)) = lim_(x->oo) (2lnx)/x 1/(1/2x^(3/2)) = lim_(x->oo) (4lnx)/x^(5/2)#
and again:
#lim_(x->oo) (4lnx)/x^(5/2) = 4 lim_(x->oo) (d/dx lnx)/(d/dx x^(5/2)) = 4 lim_(x->oo) 1/x 1/(5/2x^(7/2)) = 8/5 lim_(x->oo) 1/(x^(9/2)) = 0#
In general we have that for every #alpha > 0#:
#lim_(x->oo) lnx/x^alpha = lim_(x->oo) (d/dxlnx)/(d/dx x^alpha) = lim_(x->oo) 1/x 1/(alphax^(alpha-1)) = 1/alpha lim_(x->oo) 1/x^alpha = 0#
so: #lnx = o(x^alpha)#
So for any #a,b > 0#:
#lim_(x->oo) (lnx)^a/x^b = lim_(x->oo) (lnx/x^(b/a))^a =(lim_(x->oo) lnx/x^(b/a))^a = 0#
