You do not really need l'Hospital's rule as:

#(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (x^(2/3) (x^(1-2/3) +x^(1/2-2/3) +x ^(1/3-2/3))) / (x^(2/3)( 1 + x^(1/4-2/3))#

#(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (cancel(x^(2/3)) (x^(1/3) +x^(-1/6) +x ^(-1/3))) / (cancel(x^(2/3))( 1 + x^(-5/12))#

#(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (x^(1/3) +x^(-1/6) +x ^(-1/3)) / ( 1 + x^(-5/12))#

so as:

#lim_(x->oo) x^(1/3) +x^(-1/6) +x ^(-1/3) = +oo#

#lim_(x->oo) 1 + x^(-5/12) = 1#

#lim_(x->oo) (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = +oo#

Anyway, as the limit is in the form #oo/oo# we can apply l'Hospital's rule:

#lim_(x->oo) (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = lim_(x->oo) (d/dx (x+x^(1/2)+x^(1/3)))/(d/dx (x^(2/3)+x^(1/4)) )#

#lim_(x->oo) (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4))= lim_(x->oo) (1+1/2x^(-1/2)+1/3x^(-1/3))/(2/3x^(-1/3)+1/4x^(-3/4)) = 1/0 = +oo#