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

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. 
