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.

Take aside the $-2$, for the moment. You can use the identity
$$
a^3-b^3=(a-b)(a^2+ab+b^2),
$$
with $a=\sqrt[3]{x^3-1}$ and $b=x$. Then
\begin{align}
\lim_{x\to\infty}(\sqrt[3]{x^3-1}-x)&=
\lim_{x\to\infty}
  \frac{
    (\sqrt[3]{x^3-1}-x)
    (\sqrt[3]{(x^3-1)^2}+x\sqrt[3]{x^3-1}+x^2)
  }{\sqrt[3]{(x^3-1)^2}+x\sqrt[3]{x^3-1}+x^2}\\
&=
\lim_{x\to\infty}\frac{x^3-1-x^3}{\sqrt[3]{(x^3-1)^2}+x\sqrt[3]{x^3-1}+x^2}
\end{align}
and you should be able to continue from here. Then put in $-2$.
