Limit of $\lim_{x \to \infty}\sqrt[3]{(x+1)^2}-\sqrt[3]{(x-1)^2}$ using nothing but L'Hopitals rule I'm trying to find a general procedure for solving any limit. I went about it by transforming any indeterminate form into $\infty / \infty$ or $0/0$ and applying L'Hopitals rule and repeating this process until a determinate form is reached.
This particular problem seems to not give me a solution using this procedure. Since the indeterminate form is $\infty - \infty$, it is transformed into $$\ln \lim_{x \to \infty} \frac{e^{\sqrt[3]{(x+1)^2}}}{e^{\sqrt[3]{(x-1)^2}}}$$ which is of the form $\infty / \infty$. But applying L'Hopitals rule doesn't seem to lead anywhere after taking the derivatives. Is this method not sufficient for these kinds of limits, or am I missing something?

We have:
\begin{eqnarray*}\sqrt[3]{(x+1)^2}-\sqrt[3]{(x-1)^2}&=&x^{\frac23}\left(\left(1+\frac1x\right)^{\frac23}-\left(1-\frac1x\right)^{\frac23}\right)\\
&=&2x^{-\frac13}\frac{\left(\left(1+\frac1x\right)^{\frac23}-\left(1-\frac1x\right)^{\frac23}\right)}{\left(\left(1+\frac1x\right)-\left(1-\frac1x\right)\right)}
\end{eqnarray*}
As $x\to \infty$ we have $$\frac{\left(\left(1+\frac1x\right)^{\frac23}-\left(1-\frac1x\right)^{\frac23}\right)}{\left(\left(1+\frac1x\right)-\left(1-\frac1x\right)\right)}\to \left.\frac{d}{dt}\right\vert_{t=1}t^{\frac23}=\frac23.$$
As $\frac43x^{-\frac13}\to 0$ as $x\to\infty$, we conclude $$\sqrt[3]{(x+1)^2}-\sqrt[3]{(x-1)^2}\to 0$$ as $x\to\infty$.
