Let $B$ be Brownian motion and for $b \in \mathbb{R}$, $\sigma > 0$, let $X_t = e^{bt + \sigma B_t}$. Investigate the existence and finiteness of the a.s. limit $\lim_{t \to \infty}X_t$ according to all possible values $b, \sigma$.
My attempt:
By the law of iterated logarithm, we know that $$\limsup_{t \to \infty} \frac{B_t}{\sqrt{2t\log\log t}} = 1$$ and $$\liminf_{t \to \infty} \frac{B_t}{\sqrt{2t\log\log t}} = -1 $$ which implies $|B_t(\omega)| \le (1+\epsilon)\sqrt{2t\log\log t}$. Therefore,
Case 1: Let $b=0$. By the properties mentioned above we have $\limsup_{t \to \infty} X_t = \infty$ and $\liminf_{t \to \infty}X_t = 0$.
Case 2: Let $b < 0$. $\liminf_{t \to \infty}X_t = 0$, however I am unsure about how to calculate $\limsup X_t$. Because $\sigma B_t \to \infty $ and $bt \to -\infty$ and I'm pretty sure $\infty - \infty$ is undefined.
The solutions say that in this case $\lim_{t \to \infty}X_t = 0$. However, I don't quite see why that is true. My reasoning is because if the limit exists then that should mean $$\lim_{t \to \infty}X_t = \limsup_{t \to \infty}X_t = \liminf_{t \to \infty}X_t.$$
Hint: the law of the iterated logarithm is overkill here; the strong law of large numbers is enough. Try writing $$bt+\sigma B_t = t \left(b + \sigma \frac{B_t}{t}\right).$$