We know that #1^r+2^r+3^r+…+n^r# can be expressed as a formula in the form of a polynomial with degree #r+1#.
So, #1^2+2^2+3^2+…+n^2=an^3+bn^2+cn+d# for all #n# with certain #a,b,c,d#. This can be solved by setting #n=0,1,2,3# to get a system of linear equations:
#{(d=0),(a+b+c+d=1),(8a+4b+2c+d=5),(27a+9b+3c+d=14):}#
Solving this, we get #a=1/3,b=1/2,c=1/6,d=0#.
Thus, #1^2+2^2+3^2+…+n^2=1/3n^3+1/2n^2+1/6n=(2n^3+3n^2+n)/6#.
Use a similar process to find that #1^3+2^3+3^3+…+n^3=(n^4+2n^3+n^2)/4# and #1^6+2^6+3^6+…+n^6=(6n^7+21n^6+21n^5-7n^3+n)/42#.
Then, the original problem becomes
#lim_(n->oo)(((2n^3+3n^2+n)/6)((n^4+2n^3+n^2)/4))/((6n^7+21n^6+21n^5-7n^3+n)/42)#
#=lim_(n->oo)(42(2n^3+3n^2+n)(n^4+2n^3+n^2))/(24(6n^7+21n^6+21n^5-7n^3+n))#
#=7/12#.
See the other solution I posted for a method involving integration but is considerably easier.