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

$$
\begin{align}
2x^2 &amp;= A^3-B^3\\
x &amp;= \sqrt{\frac{A^3-B^3}{2}}
\end{align}
$$

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?
Note that $\sqrt[3]{x^3+x^2+1}=x\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^3}}$, with a similar expression for the other one. Make the substitution $t=1/x$. Our expression becomes 
$$\frac{2-\sqrt[3]{1+t+t^3}-\sqrt[3]{1-t+t^3}}{t},$$
and we want to find the limit of this as $t$ approaches $0$ from the right. Note that the conditions for using L'Hospital's Rule are met. So our limit is 
$$\lim_{t\to 0^+} -\frac{1}{3}(1+3t^2)(1+t+t^3)^{-2/3}-\frac{1}{3}(-1+3t^2)(1-t+t^3)^{-2/3}.$$
This one is easy.
As $x \to\infty$, $A\to x$ and $B\to x$. Therefore the expression would tend to $0$.