Evaluate: $\lim_{x \to \infty} \,\, \sqrt[3]{x^3-1} - x - 2$ 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.

Hint: It is best to use series. However, we can do it with algebraic manipulation. Let $a=(x^3-1)^{1/3}$ and let $b=x+2$. Multiply top and (missing) bottom by $a^2+ab+b^2$.   
Another way: Make the substitution $x=1/t$. We end up wanting
$$\lim_{t\to 0^+} \frac{\sqrt[3]{1-t^3} -1-2t}{t}.$$
Now one Hospital round does it. 
