How can I calculate $$ \lim_{(x,y)\to (1,-1)} \frac{x+y}{1+xy}$$
I want to show that this function has a maximum value of 1. I found the maxima are at (±1,±1) and for$ (1,1)$ and $(-1,-1)$ , $f(x,y)= 1$ . I am stuck on the rest two.$$$$ Edit: I will explain a little more about the situation, in special relativity, velocity addition has the form $$ \overline{v} = \frac{w+V}{1+\frac{wV}{c^2}} $$ where velocities have usual meanings. So, when a frame is moving away from you with a speed of $c$ and light flashes towards you in the opposite direction, you should get $$ \overline{v} = -c $$ because the speed of light is same in all the frame of references. How can we justify that from this equation? Essentially I have to prove that $ \overline{v}$ never exceeds $c$
We have that for $x=1+t$ and $y=-1+t$ with $t\to 0$
$$\lim_{(x,y)\to (1,-1)} \frac{x+y}{1+xy}=\lim_{t\to 0} \frac{2t}{t^2}=0=\lim_{t\to 0} \frac{2}{t}=\pm\infty$$
Added after editing
For the problem you are referring to the limit to look should be
$$\lim_{(x,y)\to (1^-,-1^+)} \frac{x+y}{1+xy}$$
since we are assuming $-1<x,y<1$.
More in general we are interested to bound the given function when $-1<x,y<1$, then we can let
and therefore by the sum identity for $\tanh$
$$\frac{x+y}{1+xy}=\frac{\tanh u+\tanh v}{1+\tanh u \tanh v}=\tanh (u+v)$$
which leads to
$$-1<\frac{x+y}{1+xy}<1$$