How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$? How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$ ? I know that if it is a square root I will multiply by the conjugate. but if it is a cubic root what shall I do?
Thanks 

Use the fact that
$$ a^3-b^3 = (a-b)(a^2+ab+b^2) \quad\Leftrightarrow\quad a-b = \frac{a^3-b^3}{a^2+ab+b^2}. $$
Then
\begin{align}
(n^2+1)^{1/3}-(n^2+n)^{1/3}
&= \frac{(n^2+1)-(n^2+n)}{(n^2+1)^{2/3}+(n^2+1)^{1/3}(n^2+n)^{1/3}+(n^2+n)^{2/3}}\\[0.2cm]
&= \frac{1-n}{(n^2+1)^{2/3}+(n^2+1)^{1/3}(n^2+n)^{1/3}+(n^2+n)^{2/3}}\\[0.2cm]
&= \frac{\frac{1}{n}-1}{\left(n^{1/2}+n^{-3/2}\right)^{2/3} + \left(n+1+\frac{1}{n}+\frac{1}{n^2}\right)^{1/3} + \left(n^{1/2}+n^{-1/2}\right)^{2/3}}\\[0.2cm]
&\to \frac{0-1}{(\infty+0)^{2/3}+(\infty+1+0+0)^{1/3}+(\infty+0)^{2/3}} = 0.
\end{align}
Wolfram confirms.
Now that I think about it, it would have probably been much simpler to use asymptotics.
