<s>[INST] How to calculate this limit with cube-roots? How to solve this limit?
$$\lim_{x\to \pm\infty}\sqrt[3]{(x-1)^2}-\sqrt[3]{(x+1)^2}$$
I figured this is the same as:
$$\lim_{x\to \pm\infty}(\sqrt[3]{x-1}+\sqrt[3]{x+1})(\sqrt[3]{x-1}-\sqrt[3]{x+1})$$
But that didn't help much I guess...
 [/INST] $$A=\sqrt[3]{(x-1)^2}-\sqrt[3]{(x+1)^2}=(x-1)^{2/3}-(x+1)^{2/3}=x^{2/3}\left( \left(1-\frac 1x \right)^{2/3}-\left(1+\frac 1x \right)^{2/3}\right)$$ Use the generalized binomial theorem or Taylor series. 
$$\left(1-\frac 1x \right)^{2/3}=1-\frac{2}{3 x}-\frac{1}{9 x^2}-\frac{4}{81
   x^3}+\cdots$$ $$\left(1+\frac 1x \right)^{2/3}=1+\frac{2}{3 x}-\frac{1}{9 x^2}+\frac{4}{81
   x^3}+\cdots$$ $$A=x^{2/3} \left(-\frac{4}{3 x}-\frac{8}{81 x^3}+\dots \right)=-\frac{4}{3 x^{1/3}}-\frac{8}{81 x^{7/3}}+\cdots$$
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