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?

This is standard $\infty-\infty$. To find $\lim f-g$ transform it to
$\lim \frac{f}{\frac{1}{1-g/f}}$. Then L'Hopital it.
