Originally Posted by

**Mario F.**
If an infinitesimal does not equal zero as you say, how come you bring it to algebra as being 0 in the final part of

your proof? It's either different or equal to 0. You have to make up your mind. Because you don't translate the meaning of the equality sign on limits into plain algebra without that resulting into you turning L into not just the limit of zero, but actually 0.

And this seems to be the problem, here, saying the meaning of the equal sign on limits is different. The equals sign never means anything other than "is equal to", even if a limit is involved:

Code:

lim[x -> infinity] 10^-x = 0

We can even put the limit in a variable:

Code:

t = lim[x -> infinity] 10^-x

And then all the following is true:

Code:

t = lim[x -> infinity] 1/x = lim[x -> infinity] -1/x
t = 0
t*5 = 0
t + 1 = 1

Because t is actually 0. Equal to. There is no special meaning to that. And one of the basic rules of equality (again, no matter if limits are involved) says:

Code:

If:
a = b
b = c
Then:
a = c

In my proof, I "showed" (the latter I didn't actually show, but those are basic identities) that:

Code:

1 - 0.9999... = lim(x -> inf) ( 1*10^-x )
lim(x -> inf) ( 1*10^-x ) = 0

And again, the equals sign for the limits has exactly the same meaning as for anything else, so we can establish from the basic rules of equality that:

Code:

1 - 0.9999... = lim(x -> inf) ( 1*10^-x ) = 0
1 - 0.9999... = 0

Why am I still arguing? Because I'm hoping I'm missing something, that I'm hoping you'll show me something I have never known. Or maybe the other way around. Though I'm still convinced I'm right.