$B(t,X)$ is the price at time $t$ of a zero coupon bond say at time $X$, time $S>T$.
When I try to compare it to the standard definition from textbooks and wikipedia I just can't make the connection.
Is $\lim_{S\rightarrow T} \frac{1}{S-T} ( \frac{B(t,T)}{B(t,S)} - 1) = -\frac{\partial}{\partial T} \ln B(t,T)$?
I have had the question unfinished in my browser for a week now, the only thinkg I can think of is $\int\frac{f'(x)}{f(x)} \, dx = \ln(f(x))$ so the left hand side should look like the ratio of a function and its derivative.
$$\lim_{S\rightarrow T} \frac{1}{S-T} \bigg( \frac{B(t,T)}{B(t,S)} - 1\bigg) = \lim_{S\rightarrow T} \frac{1}{S-T} \bigg( \frac{B(t,T) - B(t,S)}{B(t,S)}\bigg)$$ $$ = \lim_{S\rightarrow T} \frac{B(t,T) - B(t,S)}{S-T} \cdot \frac{1}{B(t,S)} =-\frac{\partial}{\partial T} \ln B(t,T)$$
OK finally I think this is there? So the first term is the derivative at $T$ ? Please let me know of any errors (or if I am way out).