You should replace the following formulas for the summations found in numerator and denominator, such that:

`1^2+2^2+...+n^2 = (n(n+1)(2n+1))/6`

`1+2+.....n = (n(n+1))/2`

Replacing the indicated formulas in the given sequence, yields:

`lim_(n->oo)(1^2+2^2+...+n^2)/(1+2+.....n) = lim_(n->oo) ((n(n+1)(2n+1))/6)/((n^2(n+1))/2)`

Reducing duplicate factors yields:

`lim_(n->oo)(1^2+2^2+...+n^2)/(1+2+.....n) = lim_(n->oo) (2n+1)/(3n)`

You need to replace `oo` for `x` in limit, such that:

`lim_(n->oo) (2n+1)/(3n) = oo/oo`

The indetermination oo/oo allows you to use l'Hospital's theorem, such that:

`lim_(n->oo) (2n+1)/(3n) = lim_(n->oo) ((2n+1)')/((3n)') `

`lim_(n->oo) ((2n+1)')/((3n)') = lim_(n->oo) 2/3 = 2/3`

**Hence, evaluating the limit of the given sequence, yields `lim_(n->oo)(1^2+2^2+...+n^2)/(1+2+.....n) = 2/3` .**