TITLE: Evaluate: $\lim_{x \to \infty} \,\, \sqrt[3]{x^3-1} - x - 2$
QUESTION [6 upvotes]: Find the following limit $$\lim_{x \to \infty} \,\, \sqrt[3]{x^3-1} - x - 2$$
How do I find this limit? If I had to guess I'd say it converges to $-2$ but the usual things like L'Hôpital or clever factorisation don't seem to work in this case.

REPLY [4 votes]: Setting $\dfrac1x=h,$
$$\lim_{x \to \infty} \,\, \sqrt[3]{x^3-1} - x - 2$$
$$=\lim_{h\to0^+}\frac{\sqrt[3]{1-h^3}-1-2h}h$$
Now using Thomas Andrews' method, $$\sqrt[3]{1-h^3}-1-2h=\frac{1-h^3-(1+2h)^3}{(1-h^3)^{\frac23}+(1-h^3)^{\frac13}(1+2h)+(1+2h)^2}$$
$$=\frac{-6h-12h^2-9h^3}{(1-h^3)^{\frac23}+(1-h^3)^{\frac13}(1+2h)+(1+2h)^2}$$
$$\implies\lim_{h\to0^+}\frac{\sqrt[3]{1-h^3}-1-2h}h$$
$$=\frac1{\lim_{h\to0^+}(1-h^3)^{\frac23}+(1-h^3)^{\frac13}(1+2h)+(1+2h)^2}\cdot\lim_{h\to0^+}\frac{(-6h-12h^2-9h^3)}h$$
$$=\frac1{(1)^{\frac23}+(1)^{\frac13}(1)+(1)^2}(-6)$$