Ans. is 3/5.
Sum of squares of first n natural numbers is n(n + 1)(2n + 1) / 6,
sum of cubes of first n natural numbers is n2(n + 1)2 / 4,
that of fourth powers is (3n2 + 3n – 1).n(n + 1)(2n + 1) / 30 and
that of fifth powers is n2(n + 1)2(2n2 + 2n – 1) / 12.
In fact you can calculate sum of kth powers of first n natural numbers if you know the sum till (k–1)th powers.

given limit is
lim
n

{n(n + 1)(2n + 1)/6}{n
2(n + 1)
2 / 4}{(3n
2 + 3n – 1).n(n + 1)(2n + 1) / 30} / {n
2(n + 1)
2(2n
2 + 2n – 1) / 12}
2Simplify above expression and get
lim
n

(2n + 1)
2 (3n
2 + 3n – 1) / 5(2n
2 + 2n – 1)
2 = (1/5) lim
n

(4n
2 + 4n + 1) (3n
2 + 3n – 1) / (2n
2 + 2n – 1)
2 = (1/5) lim
n

[(4n
2 + 4n + 1) / (2n
2 + 2n – 1)] x [lim
n

(3n
2 + 3n – 1) / (2n
2 + 2n – 1)]
= (1/5) . (4/2) . (3/2)
= 3/5.