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.