<s>[INST] Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=2-2=0$$
But on the right hand I have inrcrased $\sqrt{x^2-2x}$ rather then deceased 
Another attempt:
$$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=\lim_{x\to \infty} \sqrt[3]{x^3(1+\frac{2}{x^2})}-\sqrt{x^2(1-\frac{2}{x})}=\\=\lim_{x\to \infty} x\sqrt[3]{(1+\frac{2}{x^2})}-x\sqrt{(1-\frac{2}{x})}=\lim_{x\to \infty} x[\sqrt[3]{(1+\frac{2}{x^2})}-\sqrt{(1-\frac{2}{x})}]$$
 [/INST] 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
$$=\lim_{x\to\infty}x\left(\left(1+O(x^{-2})\right)-\left(1-\frac12\cdot\frac2x+O(x^{-2})\right)\right)$$
$$=\lim_{x\to\infty}x\left(1-1+\frac12\cdot\frac2x\right)=1$$
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