right left continuity of $f(x)=\frac{1}{x}$ at $0$ 
Determine whether $f(x)=\frac{1}{x}$ is continuous or discontinuous from the right or left at $x_0=0$

We have an asymptote at $x=0$, with $D_f=(-\infty,0) \cup (0,\infty)$. 
It follows that an equality between $f(0^-)$ and $f(0)$ is not possible. Likewise, with $f(0^+)$ and $f(0)$.
Therefore, it is discontinuous at $x_0=0$.  
Is the argumentation above correct? Is $f$ said to be "non existing" or "not defined" at $x_0$? 
Much appreciated.
 A: You don't need to analyze the limit at all.
Going through the definition of continuity:
A function $f$ is continuous at a point $c$ iff the following $3$ conditions are met:
$1)$ $f(c)$ is defined
You don't need to go any further. $f(0)$ is not defined, therefore, $f$ is not continuous at $0$.
A: $$\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{1}{x}$$
In order for a function to be continuous
$$f(a)=\lim_{x \rightarrow a}f(x)$$
For a limit of a function to exist the left hand limit must be equal to the right hand limit such that they are all equal
$$\lim_{x\rightarrow a}f(x)=\lim_{x \rightarrow a^+}f(x)=\lim_{x \rightarrow a^-}f(x)$$
So we need to prove that 
$$\lim_{x\rightarrow 0^+}f(x)=\lim_{x\rightarrow 0^+}\frac{1}{x}$$
$$\lim_{x\rightarrow 0^+}\frac{1}{x}=+\infty$$
$$\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^-}\frac{1}{x}$$
$$\lim_{x\rightarrow 0^-}\frac{1}{x}=-\infty$$
This is known as infinite discontinuity!
Since 
The function is discontinuous because
$f(0)$ is not defined at all
and 
$$\lim_{x\rightarrow 0^-}\frac{1}{x} \neq \lim_{x\rightarrow 0^+}\frac{1}{x}$$
We say that the limit does not exist at $x=0$ because left hand limit does not equal to right hand limit.
Limit at $x=0$ does not exist.
