$\varlimsup_{t\rightarrow\infty} \frac{B_t}{\sqrt{t}}>0$

I am trying to prove the following statement about the standard Brownian Motion: $\varlimsup_{t\rightarrow\infty} \frac{B_t}{\sqrt{t}}>0$. I know that it is trivial to prove the above statement by using powerful theorems about the Brownian Motion. But I have found this exercise in the first section of the first chapter in Revuz Yor. At this point, there aren't very many results about BM beyond the invariance properties. So can anyone help me to prove this using only elementary results?

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Just wondering, what is $B_t$? Thanks. –  Emmad Kareem Jan 1 '12 at 21:37
$B_{t}$ is a standard linear Brownian Motion. –  Peter Moor Jan 1 '12 at 21:39

Otherwise the limsup is $\leqslant0$ almost surely (this is an asymptotic event) and, by symmetry, the liminf is $\geqslant0$ almost surely (idem), that is, the limit exists and is $0$ almost surely. But the random variables $(B_{2^{n+1}}-B_{2^n})/\sqrt{2^n}$ are i.i.d. and unbounded, hence almost surely infinitely many of them are $\geqslant1$. This is a contradiction: if a real valued deterministic sequence $(x_n)_n$ is such that $x_n\to0$, then $\sqrt2x_{n+1}-x_n\to0$ hence $\sqrt2x_{n+1}-x_n\geqslant1$ at most finitely many times.