TITLE: Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$
QUESTION [6 upvotes]: $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
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:
$$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=\lim_{x\to \infty} \sqrt[3]{x^3(1+\frac{2}{x^2})}-\sqrt{x^2(1-\frac{2}{x})}=\\=\lim_{x\to \infty} x\sqrt[3]{(1+\frac{2}{x^2})}-x\sqrt{(1-\frac{2}{x})}=\lim_{x\to \infty} x[\sqrt[3]{(1+\frac{2}{x^2})}-\sqrt{(1-\frac{2}{x})}]$$

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