Allow us to diverge from the rules of math taught in schools and universities regarding infinities to consider this:
If I have a "number" (integer-like) that goes $$\cdots2222222222222222222222.0$$ "infinitely" to the left of the decimal point and I divide it by a number that goes $$\cdots1111111111111111111111.0$$ "infinitely" to the left of the decimal point, then I should get $2$, right?
These numbers could be rephrased as $A_\infty$ and $B_\infty$ where $A_n=\sum\limits_{i=0}^n2\cdot10^i$ and $B_n=\sum\limits_{i=0}^n1\cdot10^i$ and ∞ is just some number x satisfying "fixed point under addition by finite real or complex number" e.g. $\infty=\infty-1=1+\infty$
if I take $\lim\limits_{x\to\infty}{\frac{A_x}{B_x}}$ I get 2. In fact I get 2 for any finite positive integer value of x.
Is there some sort of "Theory of infinite-digited integers"?
UPDATE: Thanks for the feedback it has been helpful, although I have one problem with the p-adic way of handling things; saying $x = 2+10x$ means $x=2+10(2+10\cdots)=\cdots 2222222.0$ but also equal $-\frac{2}{9}$ except that I think Feynman and others have overlooked something
we have a $\phi=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\cdots}}}}}}$ which is (mostly) equal to saying $\phi=1+\frac{1}{\phi}$. The real problem comes when $x=1+\frac{1}{x}$ has more than one known solution! one being $\frac{1+\sqrt{5}}{2}$ and other being $-\frac{1+\sqrt{5}}{2}+1$ of course this specific case is just related to the square-root having 2 values for every nonzero complex number $x$, but it gets worse. I think that, $x=2+10x$ has at least 2 solutions, one being $-2/9$ and the other being the $\cdots2222222222.0$ number. The reasons why to keep them separate? Like how phi and its "counterpart" number both satisfy $x=1+\frac{1}{x}$, but they fail to match at other things and are not equal. For one thing, $\cdots22222222222.0$ number is a positive, "infinite" number (solving a sum with it as the number of iterations would turn the sum into a series) whereas $-\frac{2}{9}$ is not.
TL;DR only problem I have, $\cdots22222222222.0\neq-\frac{2}{9}$ because one is positive, infinite, and integer-y and the other is a negative fraction, though they are both counterparts for $x=2+10x$
Thanks for reading my input!
Yes, there is a theory of such numbers; these numbers are called 10-adic numbers, denoted $\mathbb{Z}_{10}$, and in the $10$-adic numbers the following manipulations are valid: we have
$$\dots 111 = \sum_{n=0}^{\infty} 10^n = \frac{1}{1 - 10} = - \frac{1}{9}$$
and
$$\dots 222 = \sum_{n=0}^{\infty} 2 \cdot 10^n = \frac{2}{1 - 10} = - \frac{2}{9}$$
and dividing them gives $2$ as expected. A related funny identity is
$$\dots 999 = \sum_{n=0}^{\infty} 9 \cdot 10^n = \frac{9}{1 - 10} = -1$$
the idea being that if you add $\dots 999$ and $1$ then you get $\dots 000$! The key feature of the $10$-adic numbers that makes all of this work is that large powers of $10$ are regarded as "small," and in particular there is a topology on the $10$-adic numbers with respect to which the series above converge.
The $10$-adic numbers have funny properties, though, the main one being that it's not true that if $ab = 0$ then either $a = 0$ or $b = 0$ (although the divisions we did above turn out to be fine). Here's a recent math.SE question where this sort of thing came up.
Via the Chinese remainder theorem $10$-adic numbers can be understood as pairs consisting of a $2$-adic number and a $5$-adic number, which are defined in the same way except that we work in base $2$ and base $5$ respectively. Because these bases are prime the resulting numbers turn out to be better behaved and mathematicians work almost exclusively with these.
A nice exercise is to show that for every prime $p$ other than $2$ or $5$ there is a $10$-adic number that deserves to be called $\frac{1}{p}$ in the sense that when you multiply by $p$ you get $1$; for example,
$$\frac{1}{7} = \dots \overline{857142}857143.$$
A harder exercise is to show that there exists a $10$-adic number that deserves to be called $\sqrt{41}$ in the sense that when you square it you get $41$ (actually there are four of them, rather than the expected two), and an even harder exercise is to determine exactly which square roots exist.