I've already used limits to argue, but it wasn't argued well. So here is a better explanation. In Calculus you learn to identify functions with holes, meaning a place in the function where continuity breaks and the function Does Not Exist. Here is an example. Take the function f(X)=((3X-2)(1-X))/(1-X). There will be a hole at X=1 because anything divided by zero will not exist. This is inarguable, it's factual that there is a hole at 1. I explicitly state this so that you know as you read the next part of this. You say that .9r=1, not the limit, not an estimate, but actual value. So, with that, let me prove you wrong. With the function f(X)=((3X-2)(1-X))/(1-X), the (1-X)s will come out to same value and divide each other out. If X=2, then f(X) will come out to be 4, because you will get ((4)(-1))/(-1). But it would be the same if you just solved 3X-2 by itself. But as stated before, the (1-X)s are there to create a hole at X=1. If you took .9r, then you would come out to a value of 1.9r. You could have an infinite amount of 9s, but no matter what you will have a value of 1.9r, but the moment that X=1, is the moment that the function will not exist. .9r=/=1, because if it did, then our function would not exist. No general algebra bullshit this time OP.