Calculate $\lim_{x \to \infty} x^{3/2}(\sqrt[3]{x^3 + 1} - \sqrt[3]{x^3 - 1})$ Calculate $$\lim_{x \to \infty} x^{3/2}(\sqrt[3]{x^3 + 1} - \sqrt[3]{x^3 - 1})$$
I found out that we can simplify it by multiplying:
$$\lim_{x \to \infty} x^{3/2}\frac {{x^3 + 1} - (x^3 - 1)} {\sqrt[3]{(x^3 + 1)^2} + \sqrt[3]{(x^3 + 1)(x^3 - 1)} + \sqrt[3]{(x^3 - 1)^2}}  =   \lim_{x \to \infty} x^{3/2}\frac {2} {\sqrt[3]{(x^3 + 1)^2} + \sqrt[3]{(x^3 + 1)(x^3 - 1)} + \sqrt[3]{(x^3 - 1)^2}}$$
So what am I supposed to do next?

Dividing both the numerator and the denominator by $x^{3/2}$ gives
$$\frac{2}{(x^{3/2}+2x^{-3/2}+x^{-9/2})^{1/3}+(x^{3/2}-x^{-9/2})^{1/3}+(x^{3/2}-2x^{-3/2}+x^{-9/2})^{1/3}}$$
