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:
$$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=\lim_{x\to \infty} \sqrt[3]{x^3(1+\frac{2}{x^2})}-\sqrt{x^2(1-\frac{2}{x})}=\\=\lim_{x\to \infty} x\sqrt[3]{(1+\frac{2}{x^2})}-x\sqrt{(1-\frac{2}{x})}=\lim_{x\to \infty} x[\sqrt[3]{(1+\frac{2}{x^2})}-\sqrt{(1-\frac{2}{x})}]$$
 A: 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 $$\lim_{x\to \infty} \left(x-\sqrt{x^2-2x}\right) = \lim_{x \to \infty} \frac{2x}{x+\sqrt{x^2-2x}} = \lim_{x\to \infty} \frac{2}{1+\sqrt{1-\frac{2}{x}}}=1$$
A: 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
$$=\lim_{x\to\infty}x\left(\left(1+O(x^{-2})\right)-\left(1-\frac12\cdot\frac2x+O(x^{-2})\right)\right)$$
$$=\lim_{x\to\infty}x\left(1-1+\frac12\cdot\frac2x\right)=1$$
A: $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}
=\lim_{x\to \infty} \frac{(\sqrt[3]{x^3+2x}^6-\sqrt{x^2-2x}^6)}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}=
=\lim_{x\to \infty} \frac{((x^3+2x)^2-(x^2-2x)^3)}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}=
=\lim_{x\to \infty} \frac{x^6+4x^4+4x^2-x^6+6x^5-12x^4+8x^3}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}=\frac{6}{6}=1$
A: Tips:
Try to find and sum up the two limits:
$$\lim_{x\to\infty} \sqrt[3]{x^3-2x}-x \text{ and }\lim_{x\to\infty}x-\sqrt{x^2-2x}$$
A: Let $\dfrac1x=h$ where $h>0$
$$\sqrt[3]{x^3+2x}= \dfrac{\sqrt[3]{1+2h^2}}h$$ and $$\sqrt{x^2-2x}=\dfrac{\sqrt{1-2h}}h$$
Now for $F=\lim_{y\to0}\dfrac{\sqrt[n]{1+my}-1}y,$  set $\sqrt[n]{1+my}-1=z,my=(1+z)^n-1$
$F=\lim_{z\to0}\dfrac{mz}{nz+O(z^2)}=\dfrac mn$
$$\implies\lim_{y\to0}\sqrt[n]{1+my}=\dfrac{my}n+1$$
