In the proportional odds model we have the the odds of survival in 1 group are proportional to the odds of survival in another group $$\dfrac{ S_1(t)}{1-S_1(t)} = \psi \dfrac{S_0(t)}{1-S_0(t)}$$
where $S_1(t)$ and $S_0(t)$ are the corresponding survival functions.
How can I show that the ratio of hazard functions $\lambda_1(t)/\lambda_0(t)\to 1$ as $t\to\infty$.
Notation: $\lambda(t) = f(t)/S(t)$ where $f(t)$ is a density function, $S(t)$ is a survival function (i.e. $S(t) = 1-F(t)$ where $F(t)$ is distribution function $f(t) = \tfrac{\partial}{\partial t}F(t)$)
We begin with an expression for $S_1(t)$
$\begin{align*} \dfrac{S_1(t)}{1-S_1(t)} &= \psi\dfrac{S_0(t)}{1-S_0(t)}\\ S_1(t)-S_1(t)S_0(t) &= \psi S_0(t)-\psi S_0(t)S_1(t)\\ S_1(t)-S_1(t)S_0(t) +\psi S_0(t)S_1(t) &= \psi S_0(t)\\ S_1(t)(1-S_0(t)+\psi S_0(t)) &= \psi S_0(t)\\ S_1(t) &= \dfrac{\psi S_0(t)}{1-S_0(t)+\psi S_0(t)} \end{align*}$
Next an expression for $f_1(t)$ $\begin{align*} f_1(t) &= \dfrac{-\partial}{\partial t} S_1(t)\\ &= \dfrac{-\partial}{\partial t}\left(\dfrac{\psi S_0(t)}{1-S_0(t)+\psi S_0(t)}\right)\\ &= \dfrac{\psi f_0(t)}{1-S_0(t)+\psi S_0(t)} + \dfrac{\psi S_0(t)(f_0(t)-\psi f_0(t))}{(1-S_0(t)+\psi S_0(t))^2}\\ &= \dfrac{\psi f_0(t)(1-S_0(t)+\psi S_0(t)) + \psi S_0(t)f_0(t) - \psi^2 S_0(t)f_0(t)}{(1-S_0(t)+\psi S_0(t))^2}\\ &= \dfrac{\psi f_0(t)-\psi S_0(t)f_0(t)+\psi^2 S_0(t)f_0(t) + \psi S_0(t)f_0(t) - \psi^2 S_0(t)f_0(t)}{(1-S_0(t)+\psi S_0(t))^2} \\ &= \dfrac{\psi f_0(t)}{(1-S_0(t)+\psi S_0(t))^2} \end{align*}$
And finally the hazard ratio
$\begin{align*} \dfrac{\lambda_1(t)}{\lambda_0(t)} &= \dfrac{f_1(t)S_0(t)}{ f_0(t)S_1(t)}\\ &= \dfrac{f_1(t)(1-S_0(t))}{\psi f_0(t)}\\ &= \dfrac{\psi f_0(t)(1-S_0(t))}{\psi f_0(t)(1-S_0(t)+\psi S_0(t))^2}\\ &= \dfrac{1-S_0(t)}{(1-S_0(t)+\psi S_0(t))^2} \to \dfrac{1-0}{(1-0+\psi 0)^2} = 1 \end{align*}$