Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$

The trick is to add and cancel an $x$ term:

$$\lim_{x\to \infty} \left(\sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\right) =\lim_{x\to \infty} \left(\sqrt[3]{x^3+2x}-x +x-\sqrt{x^2-2x}\right)$$

and compute each limit separately:

$$\lim_{x\to \infty} \left(\sqrt[3]{x^3+2x}-x\right) = \lim_{x\to \infty} \frac{2x}{\sqrt[3]{(x^3+2x)^2}+x\sqrt[3]{x^3+2x}+x^2} = 0$$

and $$\lim_{x\to \infty} \left(x-\sqrt{x^2-2x}\right) = \lim_{x \to \infty} \frac{2x}{x+\sqrt{x^2-2x}} = \lim_{x\to \infty} \frac{2}{1+\sqrt{1-\frac{2}{x}}}=1$$

Moving factors of $x^3$ and $x^2$ out, we transform the limit to $$\lim_{x\to\infty}x\left(\sqrt[3]{1+2/x^2}-\sqrt{1-2/x}\right)$$ Then applying the binomial series yields $$=\lim_{x\to\infty}x\left(\left(1+O(x^{-2})\right)-\left(1-\frac12\cdot\frac2x+O(x^{-2})\right)\right)$$ $$=\lim_{x\to\infty}x\left(1-1+\frac12\cdot\frac2x\right)=1$$

$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x} =\lim_{x\to \infty} \frac{(\sqrt[3]{x^3+2x}^6-\sqrt{x^2-2x}^6)}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}= =\lim_{x\to \infty} \frac{((x^3+2x)^2-(x^2-2x)^3)}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}= =\lim_{x\to \infty} \frac{x^6+4x^4+4x^2-x^6+6x^5-12x^4+8x^3}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}=\frac{6}{6}=1$