<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] Tips:
Try to find and sum up the two limits:
$$\lim_{x\to\infty} \sqrt[3]{x^3-2x}-x \text{ and }\lim_{x\to\infty}x-\sqrt{x^2-2x}$$
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