is $0.\overline{99}$ the same as $\lim_{x \to 1} x$? 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? 
$$
f(0.\overline{999}) = ?
$$
 A: 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. 
A: 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.
A: 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 $f(0.\overline9)=f(1)$.
A: 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. 
