The limit of $\frac{1}{x}$, as $x \to 0$ doesnt exist, or does it? Obviously, if you approach $0$ from left you get $-\infty$, if you approach from right you get $+\infty$. Ergo, the limit doesnt exist.
But what if we work in the number system of where, to real numbers we adjoin a single unsigned infinity? In that case the limit from left and right is the same: infinity. So the limit does exist. Am I missing something?
 A: Kind of. What you're describing is the one-point compactification of $\mathbb{R}$. By 'joining the ends' at a single new point $\infty$ you obtain something that looks like a circle, on which the map $x \mapsto \dfrac{1}{x}$ well-defined (and continuous) even at $0$.
The notion of 'limit' in this space makes slightly less sense, though, because there is no extension of the usual metric on $\mathbb{R}$ to a metric on $\mathbb{R} \cup \{ \infty \}$: if there were then we'd need every $x \in \mathbb{R}$ to have infinite distance from $\infty$. We can make sense of limits, but not using the $\varepsilon$-$\delta$ definition.
To be precise: open sets in $\mathbb{R} \cup \{ \infty \}$ are


*

*open sets in $\mathbb{R}$; and

*sets containing $\infty$ whose complements are closed, bounded subsets of $\mathbb{R}$


We then say $\displaystyle \lim_{x \to a} f(x) = L$ if for every open set $U \subseteq \mathbb{R} \cup \{ \infty \}$ with $L \in U$ there exists an open set $V \subseteq \mathbb{R} \cup \{ \infty \}$ containing $a$ such that, whenever $x \in V$ and $x \ne a$ then $f(x) \in U$.
According to this definition we have
$$\displaystyle \lim_{x \to 0} \dfrac{1}{x} = \infty$$
...but don't make the mistake of thinking that this implies anything about the value of $\lim_{x \to 0} \dfrac{1}{x}$ in $\mathbb{R}$: it simply doesn't exist.
