Find $\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$ without L'Hopital or Taylor series. 
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$$

My try is as follows:
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=$$$$ \lim_{x\to 
∞}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$$$=\lim_{x\to ∞}x\lim_{x\to 
∞}\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$
which is $∞×0$ , but clearly this zero is not exactly zero. I was thinking about generalized binomial 
theorem, but seems it will make the limit difficult, so how this kind of limits can be solved without using Taylor series or L'Hopital's rule?

A quite elementary way is just using the two binomial formulas $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$ as follows:
\begin{eqnarray*}\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x}
& = & (\sqrt[3]{x^{3}+3x^{2}} - x) + (x -\sqrt{x^{2}-2x})\\
& = & \frac{3x^2}{\sqrt[3]{(x^{3}+3x^{2})^2} + x\sqrt[3]{x^{3}+3x^{2}} + x^2} + \frac{2x}{x+\sqrt{x^{2}-2x}}\\
& = & \frac{3}{\sqrt[3]{(1+\frac{3}{x})^2} + \sqrt[3]{1+\frac{3}{x}} + 1} + \frac{2}{1+\sqrt{1-\frac{2}{x}}} \\
& \stackrel{x\to \infty}{\longrightarrow} & 1+1 = 2
\end{eqnarray*}
