What is $\frac{1}{\infty}$? Ok. When I say "infinity," I mean an infinitely large number (such as $9999999\ldots$)

So we all know the common proof for $0.999\ldots = 1$.

If not, here it is: 

$$
\begin{align*}
x &amp;= 0.999\ldots \\
10x &amp;= 9.999\ldots \\
10x-x &amp;= 9x \\
9.999\ldots - 0.999\ldots &amp;= 9
\end{align*}
$$

Therefore, $9x = 9, x = 1$.

Using this logic, it can be proved that $0.000\ldots1 = 0$ because 

$$
\begin{align*}
1 - 1 &amp;= 0 \\
1 - 0.999\ldots &amp;= 0.000\ldots1
\end{align*}
$$

and since $1 = 0.999\ldots$, that means that $0 = 0.000\ldots1$.

Now, if we were to take any number and divide it by an infinitely large number, then the answer would eventually consist of $0.000000\ldots1$ (or at least have an infinite series of zeros before a number other than one). Since $0.000\ldots1 = 0$, this must mean that $\frac{1}{\infty} = 0$. 

Is this correct?
(Even though the mathematics is incorrect, +1 for a well-asked question!)

Your core misunderstanding seems to be the definition of $0.999\dots$. By definition, that number equals $\frac9{10} + \frac9{10^2} + \frac9{10^3} + \cdots$, an infinite series (which equals $1$, by the proof you gave). But my point is that it has a specific mathematical definition; it's not just "notation for an idea".

On the other hand, you seem to be thinking of $0.000\dots01$ as an infinite string of $0$s followed by a $1$. Such an object does not have any mathematical meaning, and so reasoning with it will not be mathematically valid. Alternatively, you might be thinking of that expression as "what happens when I take $0.000\dots01$, with a finite number (call it $n$) of $0$s, and let $n$ get larger and larger". While I wouldn't use that notation for this idea, the idea itself is perfectly valid and leads to the true mathematical fact
$$
\lim_{n\to\infty} \frac1{10^n} = 0.
$$
In either case, your equation $1 - 0.999\dots = 0.000\dots1$ is not correct, because the number $0.999\dots$ does not have any last $9$ that would yield a $1$ after the subtraction.
We don't know, and it's depending on quite a lot of factors, too. Like, what infinity are you using? Bear in mind that $\lim\limits_{n\to \infty}n$ results into $\infty$ but $\lim\limits_{n \to \infty}n^2$ does not equal  the vanilla $n$-variant, and both get dirrefent answers. Also, you cannot get rational results using infinities. We can get close, though, using limits:
$$\lim\limits_{x \to \infty} \frac{1}{x} = 0$$
That basicly means that if we get $x$ reeaally big, $\frac{1}{x}$ will approach 0. We can say it won't actually reach 0 though, not even when $x = \infty$. Otherwise we'd get $\frac{1}{\infty} = 0$ resulting in both $1 = \infty * 0$ and $\frac{1}{0} = \infty$, which both are incorrect statements (the first one because $a = a * 1 = a * 0 * \infty = 0 * \infty = 1$, which is incorrect, and the second one because we don't divide by $0$). Furthermore, the "mistake" made by the proof that $0.999... = 1$ is that they don't use limits (for further explanation on that, look at the question I asked earlier on my profile, lots of answers there), and, you should keep in mind that you don't know if $\infty = 10^x$, so $\frac{1}{\infty}$ does not necessarily have to be $0.000...01$. That should answer most of your questions.