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So we had an interesting discussion the other day about 0.999... repeated to infinity, actually being equal to one. I understand the proof, but I'm wondering then if you had the function...

$$ f(x) = x* \frac{(x-1)}{(x-1)} $$

so $$ f(1) = NaN $$

and $$ \lim_{x \to 1} f(x) = 1 $$ what would the following be equal to?

$$ f(0.\overline{999}) = ? $$

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    Since $0.\overline{999}=1$, $f(0.\overline{999})=f(1)=NaN$.2012-11-14
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    This has nothing to do with infinitesimals.2012-11-14
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    @tbischel: do you believe that if $x = y$, then $f(x) = f(y)$?2012-11-15
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    @QiaochuYuan: I disagree with the closing of this question. The OP is not asking why $0.\overline{9} = 1$, but he/she is confused about the function $f$ and how that plays into it. I agree that the answer basically comes down to the mentioned identity, but the question is IMO now a duplicate.2012-11-15
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    You must realize that $0.\overline{999}$ is a *number*. It is one fixed number, it's not a process, not a representation for a sequence of numbers, etc. Yes the way to interpret $0.\overline{999}$ may be given by a limit, but once you write down an expression for a limit (like "let $L = \lim \dots$"), you are considering the limit $L$ as one fixed number. And of course, it need not be true that $f(L) = \lim f(\dots)$, as you can see by your own example here. If $L = 1$, then $f(L) = f(1)$ by definition.2012-11-15

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