A proof? Okay, here goes. Define f(n) as a number with the decimal representation of "0." followed by n "9"s. The value of 0.9999... with infinitely many 9s is the value of f(n) as n goes to infinity, and I'll just call it x for now. For any number epsilon, we can find an integer n0 such that, for all values of n, n >= n0, the absolute value of f(n) - x is smaller than epsilon. That's what we mean by the value as n goes to infinity.
Now look at what happens if we assume x == 1. Then (if I've done this right) we can take n0 to be the negative of the base-10 algorithm (minimum zero) plus 1, rounded up, and that works. For any other value of x, we can find an epsilon (like (1 - x)/2) where we can find a n greater than any given n0 such that the absolute value of f(n) - x is greater than epsilon.
As far as 1/infinity, 2/infinity, etc., go, you may be right if you're using infinity to represent some number too large for your number representation method. However, given two transfinite numbers of the same size, infinity1/infinity2 is completely indeterminate, since any nonzero* number times infinity is infinity. Infinity divided by infinity is 2 because 2 times infinity is infinity, and it's 3 for the exact same reason,
*It's getting late, and I'm not going to go through the zero times infinity shtick right now.