I have the equation $F=\cfrac{\beta (s-1)-(\alpha s-\beta) e^{\lambda t}}{\alpha (s-1)-(\alpha s-\beta) e^{\lambda t}}.$
such that $\lambda = \beta - \alpha$
I need to take the limit as $\alpha$ and $\beta$ tend to 1. Clearly, just trying to see what $F$ is at the limit gives $\frac{0}{0}$.
I have tried using L'Hopital's rule doing a partial derivative w.r.t $\alpha$ and this yields $\frac{st-t+s}{st-t+1}$ but I'm not sure if I can use L'hopital like this or if this a correct answer for what I'm looking for?
The limit has two possibilities.
Case 1 : $\lambda t\neq 0$
In this case, we must have
$\lim_{(\alpha,\beta)\to(1,1)} \cfrac{\beta (s-1)-(\alpha s-\beta) e^{\lambda t}}{\alpha (s-1)-(\alpha s-\beta) e^{\lambda t}}$
$= \cfrac{(s-1)-(s-1) e^{\lambda t}}{(s-1)-(s-1) e^{\lambda t}}$
$= \cfrac{(s-1)(1- e^{\lambda t})}{(s-1)(1-e^{\lambda t})}$
$=1 $ (as $e^{\lambda t} \neq 1$)
Case 2 : $\lambda t=0$
In this case, we get
$\lim_{(\alpha,\beta)\to(1,1)} \cfrac{\beta (s-1)-(\alpha s-\beta) e^{\lambda t}}{\alpha (s-1)-(\alpha s-\beta) e^{\lambda t}}$
$=\lim_{(\alpha,\beta)\to(1,1)} \cfrac{\beta (s-1)-(\alpha s-\beta)}{\alpha (s-1)-(\alpha s-\beta)}$
$=\lim_{(\alpha,\beta)\to(1,1)} \cfrac{(\beta -\alpha) s}{(\beta -\alpha)}$
$=s$