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...
 A: Hint :
Multiply expression by :
$$\frac{(\sqrt[3]{x-1})^4 + (\sqrt[3]{x-1}\sqrt[3]{x+1})^2 +(\sqrt[3]{x+1})^4 }{(\sqrt[3]{x-1})^4 + (\sqrt[3]{x-1}\sqrt[3]{x+1})^2 +(\sqrt[3]{x+1})^4}$$
After combining terms in numerator divide both numerator and denominator by $x^{4/3}$ . Result should be zero .
A: $$A=\sqrt[3]{(x-1)^2}-\sqrt[3]{(x+1)^2}=(x-1)^{2/3}-(x+1)^{2/3}=x^{2/3}\left( \left(1-\frac 1x \right)^{2/3}-\left(1+\frac 1x \right)^{2/3}\right)$$ Use the generalized binomial theorem or Taylor series. 
$$\left(1-\frac 1x \right)^{2/3}=1-\frac{2}{3 x}-\frac{1}{9 x^2}-\frac{4}{81
   x^3}+\cdots$$ $$\left(1+\frac 1x \right)^{2/3}=1+\frac{2}{3 x}-\frac{1}{9 x^2}+\frac{4}{81
   x^3}+\cdots$$ $$A=x^{2/3} \left(-\frac{4}{3 x}-\frac{8}{81 x^3}+\dots \right)=-\frac{4}{3 x^{1/3}}-\frac{8}{81 x^{7/3}}+\cdots$$
A: If $x\to+\infty,$ choose $1/x=h$
$$F=\lim_{h\to0^+}\dfrac{(1-h)^{2/3}-(1+h)^{2/3}}{h^{2/3}}$$
Method$\#1:$
$$F=\lim_{h\to0^+}\dfrac{(1-h)^2-(1+h)^2}{h^{2/3}}\cdot\dfrac1{\lim_{h\to0^+}\{(1-h)^{4/3}+(1-h)^{2/3}(1+h)^{2/3}+(1+h)^{4/3}\}}$$
Method$\#2:$
Use Taylor's Expansion for $$(1\pm h)^{2/3}=1+\dfrac23(\pm h)+\dfrac{2/3(2/3-1)}{2!}(\pm h)^2+\cdots$$
For $x\to-\infty,$ choose $-1/x=h$
