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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 $0.\overline{999}=1$, $f(0.\overline{999})=f(1)=NaN$. – Brian M. Scott Nov 14 '12 at 23:37
This has nothing to do with infinitesimals. – Asaf Karagila Nov 14 '12 at 23:38
@tbischel: do you believe that if $x = y$, then $f(x) = f(y)$? – Qiaochu Yuan Nov 15 '12 at 0:22
@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. – Thomas Nov 15 '12 at 0:28
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. – ShreevatsaR Nov 15 '12 at 6:40

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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Its just $0./bar9$ to me seems more like an alternative expression of a limit than a representation of 1. my brain is looking for a way to express it =P – tbischel Nov 14 '12 at 23:41
@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. – Asaf Karagila Nov 14 '12 at 23:44

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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$\lim_{x \to 1} f(x)$ is not equal to $f(\lim_{x \to 1} x)$ unless $f$ is continuous at $1$. – Qiaochu Yuan Nov 15 '12 at 0:50
@QiaochuYuan: And I wrote no such thing. The limits that I have are defined. – Thomas Nov 15 '12 at 0:51
@Thomas: okay, but now I don't understand why you wrote $\lim_{x \to 1} x = \lim_{x \to 1} f(x)$ (I assumed this was a typo). It's true that the OP defined the latter to be equal to $1$ but this has nothing to do with the fact that $\lim_{x \to 1} x$ is also equal to $1$, and it's also not the quantity the OP was asking about, which is $f(\lim_{x \to 1} x)$. – Qiaochu Yuan Nov 15 '12 at 0:55
@QiaochuYuan: The reason is because of the question in the title of the question. I was just trying to answer that. – Thomas Nov 15 '12 at 0:56
@Thomas: okay, but the question in the title also has nothing to do with $f$. – Qiaochu Yuan Nov 15 '12 at 0:56

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