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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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    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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By definition of what decimal notation means, $ 0.\overline 9=\sum_{k=1}^\infty9\times10^{-k}=9\sum_{k=1}^\infty10^{-k}. $ Now, using the basic formula for geometric series, $ 0.\overline 9=9\sum_{k=1}^\infty10^{-k}=9\,\frac{10^{-1}}{1-10^{-1}}=9\,\frac{\frac1{10}}{1-\frac1{10}}=9\,\frac{1}{10-1}=1. $

So, addressing your original question, your $f$ is not defined at $1$, so $f(0.\overline 9)$ makes no sense.

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While it's true that $0.\overline{9}=\lim_{x\to 1} x=1$, in your case you may not move the limit inside $f$ to get $\lim_{x\to 1}f(x)=f(1)$ since the RHS is undefined. In general, you may move the a limit from outside the function to inside the function only if the function in question is continuous at the point where you are taking the limit.

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If $0.\overline9=1$ then $f(0.\overline9)$ is as undefined as $f(1)$ is. However indeed $\lim_{x\to 1}f(x)=1$ as you said.

The reason for the above is simple. If $a$ and $b$ are two terms, and $a=b$ then $f(a)=f(b)$, regardless to what $f$ is or what are the actual terms. Once you agreed that $0.\overline9=1$ we have to have $f(0.\overline9)=f(1)$.

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    @tbischel: The fact that $0.\overline9=\lim_{n\to\infty}\frac{10^n-1}{10^n}$ does not mean that these are *different* number. No, equality means that these are just two ways to write the same thing. Similarly I could tell you to "come to my office" or I could give you the exact room number. Both things describe the same place, even though these are different phrases.2012-11-14
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The answer is that $0.\overline{9} = 1.$ So when you want to find $f(0.\overline{9})$ then that is the exact same as writing $f(1)$ which is not defined.

About the function $f$: As we have just noted, $f(1)$ is not defined. However, as you point out $\lim_{x\to 1} f(x)$ is defined and is equal to $1$. If you are interested, the fact that $f$ is not defined means that $f$ is not continuous at $1$.

To answer specifically the question in the title, you indeed have that, $0.\overline{9} = \lim_{x\to 1} x$

Sidenote: When you write $0.\overline{999}$ you can just write $0.\overline{9}$. They are the same thing.

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    @Thomas: yes, it's fine now.2012-12-17