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.
 A: We have using Taylor series
$$(x^3-1)^{\frac13}-x-2=x\left(1-\frac1{x^3}\right)^{\frac13}-x-2\sim_\infty x\left(1-\frac1{3x^3}\right)-x-2\\=-\frac1{3x^2}-2\xrightarrow{x\to\infty}-2$$
A: A simple approach, without Taylor series, is to prove that $\sqrt[3]{x^3-1}-x\to 0$ by using $$ u-v = \frac{u^3-v^3}{u^2+uv+v^2}$$ with $u=\sqrt[3]{x^3-1}$ and $v=x$. Then use that, for $x>1,\sqrt[3]{x^3-1}>x-1$, so $$|\sqrt[3]{x^3-1}-x|<\frac{1}{(x-1)^2+(x-1)x+x^2}$$
Now the right hand side clearly goes to zero as $x\to\infty$.
A: 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, $$\sqrt[3]{1-h^3}-1-2h=\frac{1-h^3-(1+2h)^3}{(1-h^3)^{\frac23}+(1-h^3)^{\frac13}(1+2h)+(1+2h)^2}$$
$$=\frac{-6h-12h^2-9h^3}{(1-h^3)^{\frac23}+(1-h^3)^{\frac13}(1+2h)+(1+2h)^2}$$
$$\implies\lim_{h\to0^+}\frac{\sqrt[3]{1-h^3}-1-2h}h$$
$$=\frac1{\lim_{h\to0^+}(1-h^3)^{\frac23}+(1-h^3)^{\frac13}(1+2h)+(1+2h)^2}\cdot\lim_{h\to0^+}\frac{(-6h-12h^2-9h^3)}h$$
$$=\frac1{(1)^{\frac23}+(1)^{\frac13}(1)+(1)^2}(-6)$$
A: 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. 
A: 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 $-2$.
A: $\displaystyle \lim_{x \to \infty} \,\, (x^3-1)^{\frac{1}{3}} - (x^{3})^{\frac{1}{3}} - 2=\lim_{x \to \infty}\frac{x^3-1-x^3}{(x^3-1)^{\frac{2}{3}}+(x^3-1)^{\frac{1}{3}}x^{\frac{1}{3}}+x^{\frac{2}{3}}}-2=-2$
A: You can do this with L'Hopital's rule and factorization. Consider
\begin{eqnarray*}
(x^3-1)^{1/3} - x - 2 & = & x \left ( \left ( 1-1/x^3 \right )^{1/3} - 1 - 2/x \right ) \\
& = & \frac{ \left (1-1/x^3 \right )^{1/3} - 1 - 2/x}{1/x}
\end{eqnarray*}
and proceed as usual.
A: We may write $$x-(x^3-1)^{1/3}=x\left[1-(1-x^{-3})^{1/3}\right]=\frac{1}{x^2}\cdot\frac{1-(1-x^{-3})^{1/3}}{x^{-3}}.$$ Rewriting in terms of $h=x^{-3}$, we observe that $$\lim_{h\to 0}\frac{1-(1-h)^{1/3}}{h}=\frac{d}{dx}(1-x)^{1/3}\Big|_{x=0}=\frac{1}{3}$$ since this is simply a difference quotient.
Hence $$\lim_{x\to\infty}\,\left[(x^3-1)^{1/3}-x+2\right]=-2-\frac{1}{3}\lim_{x\to\infty}{\frac{1}{x^2}}=-2.$$
