If you're talking about the textbook problem where it gives you the function f(x)= 0 for 2<x<3 , then I think it is possible, but I'm really not sure. Technically, if you multiply the top and bottom of 0/0 by (x-3), then you would get 0(x-3)/0(x-3), which would show that the denominator is a factor of the numerator. You could then "remove all common factors" by canceling out the (x-3)s to get the "new" function g(x)=0, and you would then plug in x=3 in the "new" function to find the value. Plugging 3 into f(x)=0 gets you 0, so the value of the extended function would be g(x)=0.
But this all seems very roundabout to me and I'm doing the homework right now, so I don't know if it's right or not. I am probably very wrong.