How can I evaluate the limit with $\alpha$ and $\beta$ tend to 1? I have the equation $F(s,t)=\cfrac{(s\alpha-\beta)e^{-\lambda t}+\beta(1-s)}{(s\alpha-\beta)e^{-\lambda t}+\alpha(1-s)}.$
such that $\lambda = \alpha - \beta$
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 $\beta$ and this yields $\frac{st-t+s}{st-t+1}$ but can I can use L'hopital like this?
 A: You have obtained the correct answer, but the reason you did was because this limit actually exists. This is a two-variable limit and checking for it's existence is less trivial than the one-variable case. To show the limit exists, we only need to prove that for the function $f(\alpha,\beta;s,t)$ defined exactly as above is independent of the angle of attack on the point $(1,1)$ in the $(\alpha,\beta)$ plane:
$$\lim_{r\to 0}f(1+r\cos\theta, 1+r\sin\theta;s,t)=H(s,t)$$
Indeed, substituting the polar coordinates and performing a standard limiting procedure reveals that
$$\lim_{r\to 0}f(1+r\cos\theta, 1+r\sin\theta;s,t)=\lim_{r\to 0}\frac{(s-1)\frac{e^{-r(\cos\theta-\sin\theta)}-1}{r}+(s\cos\theta-\sin\theta)e^{-r(\cos\theta-\sin\theta)}+\sin\theta(1-s)}{(s-1)\frac{e^{-r(\cos\theta-\sin\theta)}-1}{r}+(s\cos\theta-\sin\theta)e^{-r(\cos\theta-\sin\theta)}+\cos\theta(1-s)}$$
All the individuals summands in numerator and denominator now have limits that exist so we can deduce after some miraculous simplification of the angle dependence,
$$H(s,t)=\frac{(1-s)t+s}{(1-s)t+1}$$
and hence the limit exists. If the limit did not exist, you would get different answers by taking L'Hopital's rule with $\partial_\alpha$ and $\partial_\beta$, for example.
Note here that this function cannot be reduced to a one variable function, but to first order in Taylor polynomials the numerator and denominator only depend on the combination $\alpha-\beta$, which makes the existence of this limit possible.
