Let $(B(t))_{t∈[0,\infty)}$ be a Brownian Motion. We want to show that $$ \frac{B(t)}{t}\to^{a.s.} 0$$ when $t\to \infty$.
For that we want to use the fact that the Brownian motion $\tilde B(t):=t B(t)$ is continuous at $0$.
The proof then says that for almost all $w \in \Omega $: $$\lim_{t \to \infty} \frac{B(t)(w)}{t}=\lim_{t \to \infty} \tilde B(\frac{1}{t})(w)=0$$
But I feel that something is wrong here. Isn't $\tilde B(\frac{1}{t})=B(\frac{1}{t})/t$ rather than $B(t)/t$?
Showing continuity of the reflected(at infinity) brownian motion is equivalent to showing that $\frac{B_{t}}{t}\to 0$.
That is, $X_{t}=tB_{\frac{1}{t}}$ is a process with same law as that of Brownian motion. But if you manage to show that $\lim_{t\to 0}X_{t}=0$ , then you have that $X_{t}$ is a Standard Brownian Motion called the projective reflection at infinity.
Continuity away from $0$ is trivial to show but, showning continuity at $0$ is the main step.
If you do that, you'll get that $\lim_{t\to 0}\frac{B_{1/t}}{1/t}\to 0$ which is same as $\lim_{t\to\infty}\frac{B_{t}}{t}\to 0$.
And conversely if you manage to show $\lim_{t\to\infty}\frac{B_{t}}{t}\to 0$ then you have that $X_{t}$ is a Brownian Motion.
See my answer here for a proof which imitates the one given in Rene Schilling.
One can use Doob's Maximal Inequality to give a direct proof of $\lim_{t\to\infty}\frac{B_{t}}{t}\to 0$ by restricting to intervals of the form $[n,n+1]$. See here or here for a proof.