`lim_(n->oo)` `root(3)((n+1)^2) -root(3)((n-1)^2)`

Hello!

We'll use an identity `a^3 - b^3 = (a - b)(a^2 + ab + b^2)` in the form `a - b = (a^3 - b^3) /(a^2 + ab + b^2)`  for  `a = root(3)((n+1)^2)`  and  `b = root(3)((n-1)^2).`

This way we obtain

`root(3)((n+1)^2) - root(3)((n-1)^2) = ((n+1)^2 - (n-1)^2) / ((n+1)^(4/3) + (n+1)^(2/3)(n-1)^(2/3)+(n-1)^(4/3)) =`

`=(4n) /((n+1)^(4/3) +(n+1)^(2/3)(n-1)^(2/3)+(n-1)^(4/3)).`

All three terms in the denominator are equivalent to `n^(4/3)` as `n->oo.` Therefore the limit is the same as  `lim_(n->oo) (4n) / (3n^(4/3)) = lim_(n->oo) (4) / (3n^(1/3))= 0.`

This is the answer (zero), and it is true for `n->+oo` and `n->-oo.`

To prove that, say,  `(n+1)^(4/3)`  is equivalent to  `n^(4/3),`  consider

`(n+1)^(4/3) / n^(4/3) = ((n+1)/n)^(4/3) = (1+1/n)^(4/3),` which tends to `1` as as `n->oo.`