Q:

How to find the limit of a function with a third root?

I have the function $x^2(\sqrt[3]{x^3+1} - x) $ and have to find the limit for $x \rightarrow \infty $.
After many hours of forming around I still have no clue how to find it. Is there anybody who could give me a hint?

A:

Hint: Use the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$  with $a=\sqrt[3]{x^3+1}$ and $b=x$ to rationalize the expression.
$$\begin{align}
x^2(\sqrt[3]{x^3+1}-x) & =x^2(\sqrt[3]{x^3+1}-x)\frac{\sqrt[3]{(x^3+1)^2}+x\sqrt[3]{x^3+1}+x^2}{\sqrt[3]{(x^3+1)^2}+x\sqrt[3]{x^3+1}+x^2}\\
& ={x^2\over \sqrt[3]{(x^3+1)^2}+x\sqrt[3]{x^3+1}+x^2}
\end{align}$$
Note that $a^3-b^3=1$ when $a=\sqrt[3]{x^3+1}$ and $b=x$.
Now, multiply top and bottom by $1/x^2$ and take the limit.

