<s>[INST] Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$ 
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$

My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now
$$
\begin{align}
2x^2 &= A^3-B^3\\
x &= \sqrt{\frac{A^3-B^3}{2}}
\end{align}
$$
So the limit becomes
$$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$
How can I complete the solution from this point?
 [/INST] It is easier to break into two limits, $x-\sqrt[3]{x^3+x^2+1}$ and $x-\sqrt[3]{x^3-x^2+1}$ then we shall see the first limits to $-\frac{1}{3}$ and the second $\frac{1}{3}$ so the sum limits to zero.
Note the more general result,
$$x-\sqrt[n]{x^n +ax^{n-1}+\cdots}\rightarrow -\frac{a}{n}$$
If we write $A=\sqrt[n]{x^n +ax^{n-1}+\cdots}$ then 
$$x-A=\frac{x^n-A^n}{x^{n-1}+x^{n-2}A +\cdots +A^{n-1}}=\frac{-ax^{n-1}+\cdots}{x^{n-1}+x^{n-2}A +\cdots +A^{n-1}}$$
and since $\frac{A}{x}\rightarrow 1$ the rest is easy. 
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