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.
 A: When you pick out $n^3$, you should get $$-n(1+\frac{1}{n^3})^{1/3}+n(1+\frac{2}{n}+\frac{4}{n^3})^{1/3}$$
The cube-root of $1+x$ is close to $1+(x/3)$ if $x$ is small, so your expression is close to 
$$-n(1+\frac{1}{3n^3})+n\left(1+\frac{2}{3n}+\frac{4}{3n^3}\right)$$
As you can see, the largest term that remains is n(2/3n) = 2/3
A: 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 $\frac{2}{1+1+1} = 2/3$.
