Mark44
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The conclusion above is not valid (which is probably the point you're trying to make).votingmachine said:First:
0x=0 for all x. I don't see any reason to prove it. I think I could wrestle thru an inductive proof for all integers, but it also seems obvious.
It seems that the loose math that was used relies on combining limits with other associative (or commutative?, algebraic?) properties. I don't know if that is valid.
EG:
(x=>infinity)limit(2x)=infinity
(x=>infinity)limit(3x)=infinity
And infinity=infinity so 2x=3x and 2=3
Just because ##\lim_{x \to \infty} f(x) = L## and ##\lim_{x \to \infty} g(x) = L##, it doesn't follow that f(x) = g(x).
Similar comment as above.votingmachine said:or
(x=>0)limit(2x)=0
(x=>0)limit(3x)=0
And 0=0 so 2x=3x and 2=3
votingmachine said:in both cases the use of loose math may well be too loose.
Infinity is one of those symbols that may be too easy to throw around.

