I saw a video on logarithms saying if there is a limit where $x$ approaches $\pm\infty$ of some fraction, then we can solve by using these rules:
If the largest power on the top and bottom are the same, then the limit is the division of the leading coefficients;
If the largest power is on the bottom, then the limit is $0$;
if the largest power is on the top then there is no horizontal asymptote.
For example, $\lim_{x\to\infty}\frac{4x+2}{2x+8}$ is $\frac{4}{2}$ or $2$ because the powers of $x$ are the same and we divide the leading coefficients.
Another one: $\lim_{x\to-\infty}\frac{x+11}{x^3+16}$ is $0$ because the largest power is on the bottom.
My question: How do I compute the limit $\lim_{x\to\infty}\frac{\log x}{x^{1/k}}$? This is particularly difficult for me because of the arbitrary variable $k$, and the fact that the largest power on top ($1$) is only greater than $\frac{1}{k}$ for all $k > 1$.
So there's a case where they can both be the same (where $k=1$ in which case we divide the leading coefficients which is 1), or the case where the largest power is on top ($k>1$), in which case there is no horizontal asymptote.
Hint: Use L'Hopital's Rule.${}$
I.e. $$\lim_{x \to \infty}\frac{\log x}{x^{1/k}}=\lim_{x \to\infty}\frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(x^{1/k})}.$$