Online I found the asymptotic behavior property of geometric Brownian Motion $X_t$as:
If $\mu$ (drift parameter) is $\ge$ $\sigma^2/2$ where $\sigma$ is the volatility parameter, then $X_t \rightarrow \infty$ as $t \rightarrow \in$
If $\mu$ is $\lt$ $\sigma^2/2$, then $X_t \rightarrow 0$ as $t \rightarrow \infty$
If $\mu$ is $=$ $\sigma^2/2$, then $X_t$ has no limit as $t \rightarrow \infty$
While this makes sense, how would the proof look like for this property? I'm not really sure how to approach it at the moment. Any help is appreciated.