I quote Delbaen and Shirakawa (2002).

Starting from a stochastic differential equation of the form: $$dr_t=\alpha\left(r_{\mu}-r_t\right)dt+\beta\sqrt{\left(r_t-r_m\right)\left(r_M-r_t\right)}dW_t\tag{1}$$ with $\left\{W_t\right\}_{t\geq0}$ a standard Wiener process in the filtered probability space $\left(\Omega,\mathcal{F},\left\{\mathcal{F}_n\right\},\mathbb{P}\right)$. We assume $\alpha,\beta>0$ and $r_m=0<r_{\mu}<r_M=1$, which guarantee the existence of stationary distribution.

Let us first consider the variable transformation: $$z_t=\dfrac{r_t-r_m}{r_M-r_m}$$ whence SDE (1) can be rewritten as: $$dz_t=\alpha(\gamma-z_t)dt+\beta\sqrt{z_t(1-z_t)}dW_t\tag{2}$$ with $\gamma=\dfrac{r_{\mu}-r_m}{r_M-r_m}$.

Let us consider $r_m=0$ as a lower bound and $r_M=1$ as an upper bound. Let $\tau_y$ be the stopping time: $$\tau_y=\inf\left\{t\geq0: z_t=y\right\}, y\in\left\{r_m=0, r_M=1\right\}$$ Then, let $\rho_{x,y}$ be the probability that $z_t$ hits $y$ in finite time when it starts from $x$. Namely: $$\rho_{x,y}=\mathbb{P}\left(\tau_y<\infty|z_0=x\right)$$Then, it holds that: $$\rho_{x,0}=\lim\limits_{y\to0, z\to0}\dfrac{B_{x,z}(p,q)}{B_{y,z}(p,q)}\tag{3}$$ $$\rho_{x,1}=\lim\limits_{y\to0, z\to0}\dfrac{B_{y,x}(p,q)}{B_{y,z}(p,q)}\tag{4}$$ where: \begin{cases} p=1-\dfrac{2\alpha\gamma}{\beta^2}\\ q=1-\dfrac{2\alpha(1-\gamma)}{\beta^2}\\ B_{x,y}(u,v)=\displaystyle{\int_x^y}z^{u-1}(1-z)^{v-1}dz \end{cases}

What I cannot really understand is the **bold** part, in particular $(3)$ and $(4)$, with $B_{x,y}(u,v)$, $p$ and $q$ defined as it immediately follows below $(3)$ and $(4)$ $\bigg($For example, in $(4)$ I would expect $\lim\limits_{y\to\color{red}{1},z\to\color{red}{1}}(\cdots)\bigg)$. Why are they defined that way? Could you please give me some explanation for such "results"?