Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^3$.

Use the following fact: when $x\to 0$,
$$(1+x)^{\alpha}-1\sim\alpha x.$$
So 
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}=n\left(\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)\right)$$
$$=\frac{\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)}{\frac{1}{n}},$$
use the above fact:
$$\sqrt[3]{1+\frac{3}{n}}-1 \sim \frac{1}{n},\sqrt{1+\frac{2}{n}}-1\sim \frac{1}{n},$$
and this implies
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )=0.$$
In fact:
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}\sim -\frac{1}{2n},n\to\infty.$$
