Intersection-exponent for one-dimensional Brownian motion

We let $$B^1,B^2$$ be independent, one-dimensional Brownian Motions with $$B^1(0)=-1$$ and $$B^2(0)=1$$ and $$T_n^i=\inf\{t\geq0:|B^i(t)|=n\}$$.

In Gregory Lawler's: Hausdorff Dimension of Cut-Points for Brownian Motion, it is claimed, that

$$\textbf{P}\{B^1[0,T_n^1]\cap B^2[0,T_n^2]=\emptyset\} \approx n^{-2}$$

where for functions $$f$$ and $$g$$, $$f\approx g$$ means $$\lim_{n\to \infty}\frac{\ln f(n)}{\ln g(n)}=1$$.

Now I think, that in this case even

$$\lim_{n\to\infty}\frac{\textbf{P}\{B^1[0,T_n^1]\cap B^2[0,T_n^2]=\emptyset\}}{n^{-2}}=1\quad$$ without the logarithms holds true.

Note that

$$\textbf{P}\{B^1[0,T_n^1]\cap B^2[0,T_n^2]=\emptyset\}=\textbf{P}\{\max\limits_{t\in[0,T_n^1]}{B^1(t)}\leq \min\limits_{t\in[0,T_n^2]}{B^2(t)}\}$$,

and we get an upper bound by the application of the Optional Stopping Theorem:

$$\textbf{P}\{\max\limits_{t\in[0,T_n^1]}{B^1(t)}\leq \min\limits_{t\in[0,T_n^2]}{B^2(t)}\}\leq \textbf{P}\{\max\limits_{t\in[0,T_n^1]}{B^1(t)}\leq 1; \min\limits_{t\in[0,T_n^2]}{B^2(t)}\leq -1\}=(\frac{1}{n+1})^2$$.

However I am struggling with finding a lower bound. I was thinking to use the distribution of the maximum-process $$M_t=\min\limits_{0\leq s\leq t}{B(s)}$$, but I was not successful so far.

Any help would be greatly appreciated!