finding the limit of a series with 3rd root this is not a homework question.
how do i reach the limit of:

i tried taking out $n^3$ and then i got $n-n$ which is $0$, but the true answer is $2/3$.
i can't understand why the answer is $2/3$ and what method to use here.
thank you very much in advance,
yaron.

Perhaps easier to simplify first. We want to get rid of the cube roots, so remember that $a^3-b^3 = (a-b) \cdot \left(a^2 + ab + b^2\right)$. Note that
$$
\begin{split}
\sqrt[3]{n^3+2n^2+4} - \sqrt[3]{n^3+1}
   &= \frac{\left(\sqrt[3]{n^3+2n^2+4} - \sqrt[3]{n^3+1}\right)
            \left( {(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \right)}
           {(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \\
   &= \frac{n^3+2n^2+4 - n^3 - 1}
           {(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \\
   &= \frac{2n^2+3}
           {(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \\
\end{split}
$$
Divide through by $n^2$ and with $n \to \infty$ this will look like $\frac{2}{1+1+1} = 2/3$.
