Finding $\limsup$ of a Brownian motion function Q) Use the fact that $(1+t)^{-1/2}\exp(B_t^2/2(1+t))$ is a martingale to show that $$\limsup_{t\to \infty} \frac{B_t}{\sqrt{(1+t)\log(1+t)}}\leq 1 \text{ a.s.}$$
I can see that $(1+t)^{-1/2}\exp(B_t^2/2(1+t))$ is a non-negative martingale and hence converges to a finite limit a.s. 
$$\begin{align}
x_t &:= \frac{B_t}{\sqrt{(1+t)\log(1+t)}} \tag{1}\\
\implies (1+t)^{-1/2}\exp(B_t^2/2(1+t)) &= (1+t)^{-1/2}\exp(x_t^2\log(1+t)) \tag{2}
\end{align}
$$
Thus if $x_t^2\geq 1/2$ i.o., then RHS of $(2)\geq 1$ i.o. which means LHS of $(2)\geq 1$ i.o. but can't the martingale have its limit as a number/random variable $>1$ so that LHS of $(2)\geq 1$ i.o. is okay? If that is right, may I know how to prove that $\limsup_{t\to \infty}x_t\leq 1$ a.s.? 

If $\lim \sup x_t^{2}  >1$ then  $(1+t)^{-1/2} e^{x_t^{2} log(1+t)} \to \infty$ along some sequence $(t_n)$ and this contradicts the fact that the martingale converges to a finite limit. 
[$(1+t_n)^{a_n-\frac 1  2} \to \infty$ if $a_n >1$ for all $n$ and $t_n \to \infty$].
