How do you find the limit #(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4))# as #x->oo#?

The greatest power of #x# here is #1#, and is in the numerator. A handy rule of thumb is that then the numerator will outgrow the denominator, so the limit is #oo#.

Another way to approach this is to factor the greatest power from the numerator and denominator:

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

#=lim_(xrarroo)(x^(1/3)(1+x^(-1/2)+x^(-2/3)))/(1+x^(-5/12))#

#=lim_(xrarroo)x^(1/3)*lim_(xrarroo)(1+x^(-1/2)+x^(-2/3))/(1+x^(-5/12))#

Note that for #p>0#, #lim_(xrarroo)x^-p=0#, so this limit becomes:

#=lim_(xrarroo)x^(1/3)*(1+0+0)/(1+0)#

#=lim_(xrarroo)x^(1/3)#

#=oo#

#lim_(xrarroo)(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4))# has initial form, #oo/oo# which is indeterminate.

We will write a quotient that is equivalent, but whose denominator does not go to infinity.

Factor out of both the top and bottom, the greatest power of #x# in the bottom. That is, factor out #x^(2/3)#.

#(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (x+x^(6/12)+x^(4/12))/(x^(8/12)+x^(3/12))#

# = (x^(8/12)(x^(1/3)+1/x^(2/12)+1/x^(4/12)))/(x^(8/12)(1+1/x^(9/12))#

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

#lim_(xrarroo)(x^(1/2)+1/x^(1/6)+1/x^(1/3))/(1+x^(3/4)) = (oo+0+0)/(1+0) = oo#