Finding the limit of a sequence $\lim _{n\to \infty} \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right)$ If there were a regular square root I would multiply the top by its adjacent and divide, but I've tried that with this problem and it doesn't work. Not sure what else to do have been stuck on it. 
$$ \lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}-
\sqrt [3]{n} \right) .$$

$$
\begin{align*}
\lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}-
\sqrt [3]{n} \right) 
&= \lim _{n\to \infty } \sqrt [3]{n^2} \cdot \sqrt[3]{n} \left( \sqrt [3]{1+ \frac{1}{n}}-
1 \right) 
\\ &= \lim _{n\to \infty } n \left( \sqrt [3]{1+ \frac{1}{n}}-
1 \right) 
\\ &= \lim _{n\to \infty } \frac{\sqrt [3]{1+ \frac{1}{n}}-
1 }{\frac{1}{n}}
\\ &= \lim _{h \to 0} \frac{\sqrt [3]{1+ h}-
1 }{h}
\\ &= \left. \frac{d}{du} \sqrt[3]{u} \ \right|_{u=1}
\\ &= \cdots
\end{align*}
$$
