# Asymptotic behavior property of geometric Brownian Motion proof

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 $$t \rightarrow \infty$$

If $$\mu < \sigma^2/2$$, then $$X_t \rightarrow 0$$ as $$t \rightarrow \infty$$

If $$\mu = \sigma^2/2$$, 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.

For any $$\alpha > 0$$, \begin{align*} \lim_{t\rightarrow\infty}P\left(e^{\big(u-\frac{\sigma^2}{2}\big) t +\sigma W_t} > \alpha \right) &= \lim_{t\rightarrow\infty}P\left(\big(u-\frac{\sigma^2}{2}\big) t +\sigma W_t > \ln \alpha \right)\\ &=\lim_{t\rightarrow\infty}P\left(\frac{W_t}{\sqrt{t}} > \frac{\ln\alpha- \big(u-\frac{\sigma^2}{2}\big) t }{\sigma \sqrt{t}} \right)\\ &=\lim_{t\rightarrow\infty}\Phi\left(\frac{\big(u-\frac{\sigma^2}{2}\big) t -\ln\alpha}{\sigma \sqrt{t}} \right)\\ &= \begin{cases} 1, &\mbox{ if } u>\frac{\sigma^2}{2},\\ \frac{1}{2}, &\mbox{ if } u=\frac{\sigma^2}{2},\\ 0, &\mbox{ if } u < \frac{\sigma^2}{2}. \end{cases} \end{align*} The conclusion now follows immediately.
Let $$X_t = e^{\big(u-\frac{\sigma^2}{2}\big) t +\sigma W_t}$$. Note that \begin{align*} \left(\omega:\, \lim_{t\rightarrow \infty} X_t = \infty \right) &= \cap_{m=1}^{\infty}\cup_{n=1}^{\infty} \cap_{t\ge n}(\omega:\,X_t > m). \end{align*} The almost surely convergence, for the case where $$u>\frac{\sigma^2}{2}$$, follows immediately.
Write $X_t = \exp(\mu B_t + (\mu - \frac {\sigma^2}{2})t )$. To prove the statement you can use the law of large numbers for Brownian motion which states that $\lim_{t \to \infty} \frac {B_t}{t} = 0$. Then rewrite $X_t$ as $$X_t = \exp(t (\mu \frac {B_t}{t} + (\mu - \frac{\sigma^2}{2})).$$ Using these two properties, you can analyze the convergence behaviour of $X_t$.