Are $\frac{\ln x_n}n\xrightarrow{n\to\infty}x$ and $x_n^{\frac1n}\xrightarrow{n\to\infty}y$ equivalent?

Question

Let $(x_n)_{n\in\mathbb N}\subseteq[0,\infty)$. Consider the following conditions:

1. There is a $x\in[-\infty,\infty)$ with $$\frac{\ln x_n}n\xrightarrow{n\to\infty}x.$$
2. There is a $y\in[0,\infty)$ with $$x_n^{\frac1n}\xrightarrow{n\to\infty}y.$$

Are these conditions equivalent? I'd say: Of course, they are! (And $y=e^x$.)

But usually I see such statement given in the form of 1., even when one is actually interested in $\ln x$ and not $x$ itself. So, is there some subtlety that I'm missing? The only thing I could imagine is $x_n=0$ or $x=-\infty$, but using $e^{-\infty}=0$ and $\ln0=-\infty$, I don't see a problem ...