Evaluate a limit using L'Hopital's Rule

As $x \to \infty$, $\ln(x) \to \infty$ and $x^{1/3} \to \infty$, so direct substitution gives the indeterminate form $\infty/\infty$. This justifies applying L'Hopital's Rule, which says we may instead compute the limit of the ratio of derivatives, provided that limit exists.

Differentiating numerator and denominator separately: the derivative of $\ln(x)$ is $1/x$, and the derivative of $x^{1/3}$ is $\frac{1}{3}x^{-2/3}$. So the limit becomes $\lim_{x\to\infty} \frac{1/x}{\frac{1}{3}x^{-2/3}} = \lim_{x\to\infty} \frac{3x^{-2/3}}{x} = \lim_{x\to\infty} 3x^{-2/3-1} = \lim_{x\to\infty} \frac{3}{x^{5/3}}.$

As $x \to \infty$, $x^{5/3} \to \infty$, so $3/x^{5/3} \to 0$. Therefore the original limit equals $0$.

No, it was not still indeterminate. After differentiating once, the expression simplified to three divided by x to the five thirds power, which as x approaches infinity is simply a finite number divided by something growing without bound. This is not of the form infinity over infinity or zero over zero, so no further application of L'Hopital's Rule was needed; the limit could be evaluated directly to get zero.