exist as in a finite number?
If that's the case then
f(x) = 1/(2-x)
g(x) = 1/(x-2)
x->2 for f(x) would be +/- infinity depending on direction
x->2 for g(x) would also be +/- infinity depending on direction
however f(x)+g(x) equals 0, so x->2 equals 0
Not sure if that's what you're looking for