I'm finishing up a course on stochastic calculus and there's a proof in my notes that isn't making sense to me. We want to show that if $a < \frac{1}{2}$, then the Bessel process $X$, defined as the solution of the SDE, $$dX_{t} = \frac{a}{X_{t}} dt + dB_{t}$$ where $B$ is a Brownian motion and $t < T = \inf \{t \geq 0 \, \mid \, X_{t} = 0\}$, hits the origin in finite time. In particular, we want to prove $\mathbb{P}^{x}\{T < \infty\} = 1$ for each $x > 0$. (Here $\mathbb{P}^{x}$ denotes the law of $X$ with $X_{0} = x$.)
What we proved in class is the following: for each $0 < r < 1 < R < \infty$, if $\varphi_{r,R}(x) = \mathbb{P}^{x}\{X \, \, \text{hits} \, \, r \, \, \text{before} \, \, R\}$, then $\varphi_{r,R}$ is given by
$$\varphi_{r,R}(x) = \frac{R^{1 - 2a} - x^{1 - 2a}}{R^{1 - 2a} - r^{1 - 2a}}.$$
Thus,
$$\mathbb{P}^{x}\{X \, \, \text{gets to zero before} \, \, R\} = \lim_{r \to 0^{+}} \varphi_{r, R}(x) = 1 - \left(\frac{x}{R}\right)^{1 - 2a}.$$
My notes claim this implies $\mathbb{P}^{x}\{T < \infty\} = 1$. That's not clear to me. (I do see that, by Borel-Cantelli, $X$ won't hit infinitely many of the integers, and, therefore, by the strong Markov property and density of the rationals, it's doomed to spiral towards the origin eventually.)
Why can't $X$ take an infinitely long time to reach zero? Certainly, we know deterministic functions whose limit at infinity is zero even though they never reach zero. In other words, $\lim_{t \to \infty} \frac{1}{t} = 0$, but it isn't true that this limit is achieved in finite time.
Any help would be much appreciated.
I think I can answer my own question so here's a shot.
First, recall that the Brownian motion $B$ almost surely hits every positive number. In fact, $\limsup_{t \to \infty} t^{-\frac{1}{2}} B_{t} = \infty$ almost surely.
Now $X$ can be written as $X_{t} = \int_{0}^{t} \frac{a}{X_{s}} \, ds + B_{t}$ and if $t < T$, then this yields $X_{t} \geq B_{t}$.
Therefore, $$\mathbb{P} \left\{T = \infty, \, \, \limsup_{t \to \infty} X_{t} = \infty \right\} = \mathbb{P}\{T = \infty\}.$$
However, I already argued that $\lim_{t \to \infty} X_{t} = 0$ almost surely. Consequently, the left-hand side is zero, which implies $\mathbb{P}\{T < \infty\} = 1$.