I am reading the following paper on a rigorous construction of Brownian Motion: Brownian Motion. In the paper, they give a peculiar proof of the fact that the time inverted Brownian Motion is continuous at $t=0$ (pg.4). Specifically, a time-inverted Brownian Motion is given to be:
$$ X_t = \begin{cases} 0 &: t=0 \\ tB_{1/t} &: t \neq 0 \end{cases} $$
They say that $X_t$ has the distribution of a Brownian Motion on $\mathbb{Q}$, the set of all rationals, so that:
$$ 0 = \lim_{n \to \infty}X(\frac{1}{n}) = \lim_{t \to 0}X(t) $$
What I understand is how they get the second equality, which is obtained via setting $n = \frac{1}{t}$. However, what is confusing me and what I dont understand is how the first equality, $0 = \lim_{n \to \infty}X(\frac{1}{n})$ is obtained. I am assuming it has something to do with real analysis but am not really sure why $0 = \lim_{n \to \infty}X(\frac{1}{n})$ is true. Could someone give me a hint? Thanks.
