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For each $t>0$, $B_t/\sqrt{t}$ has a standard normal distribution. Therefore for any $k$ we have $\mathbb{P}(B_t/\sqrt{t} \ge k) = 1-\Phi(k) > 0$ where $\Phi$ is the standard normal cdf. So if we fix a sequence $t_n \downarrow 0$ and let $X = \limsup_{t_n \downarrow 0} B_{t_n}/\sqrt{t_n}$ we have $\mathbb{P}(X \ge k) \ge 1-\Phi(k) > 0$. But by the Blumenthal 0-1 law, $X$ is almost surely constant, so $X \ge k$ almost surely. Now $k$ was arbitrary so $X = +\infty$ almost surely. In particular, almost surely, $\sup_{t > 0} B_t/\sqrt{t} = +\infty$.