Default intensity in Black-Cox model

Consider the model by Black and Cox (Journal of Finance, 1976).

The default intensity function is defined in the usual way: $$h(t) \equiv - \frac{\partial \log P[\tau > t| \mathcal{F}_t]}{\partial t}$$ where $$\tau$$ is the first hitting time of a constant absorbing barrier $$V_b$$ of a Geometric Brownian motion $$V_t$$, and $$\mathcal{F}_t$$ is the filtration up to time $$t$$. In the Black-Cox model, $$h(t) \in \{0, \infty\}$$. How can one prove this?

As shown in Credit Risk Modeling Notes (Bielecki, Jeanblanc, Rutkowski), Corollary 1.3.1, for $$t < s$$, we have:

$$P(\tau \leq s | {\cal F}_t) = N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right ) + {\rm e}^{-2\nu \sigma^{-2}Y_t} N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right ),$$

where

$$Y_t = y_0+ \nu t +\sigma W_t, \: \sigma >0,$$ $$\tau = \inf \; \{t\geq 0 | Y_t = 0 \},$$ and $$N$$ is the standard normal cdf.

(In your notations, $$Y_t$$ is the distance to default, $$Y_t =\ln (V_t/V_b)$$.)

We then calculate the conditional density probability as follows: $$\frac{\partial P(\tau \leq s | {\cal F}_t)}{\partial s}$$ $$= n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right) \left( 2^{-1}Y_t \sigma^{-1}(s-t)^{-3/2}- 2^{-1}\nu(s-t)^{-1/2}\right)$$ $$+ {\rm e}^{-2\nu \sigma^{-2}Y_t} n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right) \left( 2^{-1}Y_t \sigma^{-1}(s-t)^{-3/2}+ 2^{-1}\nu(s-t)^{-1/2}\right)$$

$$= n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right) Y_t \sigma^{-1}(s-t)^{-3/2},$$

noting that $${\rm e}^{-2\nu \sigma^{-2}Y_t} n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right) = n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}-\nu(s-t)^{1/2}\right),$$

where $$n$$ is the standard normal pdf.

Using L'Hospital we get:

$$\lim_{s\rightarrow t^+} \frac{\partial P(\tau \leq s | {\cal F}_t)}{\partial s} =0.$$