Prove how this function is discontinuous I am trying to prove why this function is discontinuous based on the three conditions

function exists at $x=a$ (In other words, $f(a)$ is a real number)
the limit of the function exists at $x=a$. (That is, $\lim_{x\to a}f(x)$ is a real number)
The two values are equal (That is, $\lim_{x\to a}f(x)=f(a)$.) 

$$f(x)\begin{cases}
 x^2   & \text{if $x<2$} \\
 3x-2  & \text{if $x>2$} \\
\end{cases}$$
at $x=2$ (I don't know how to put this to the right of the branches, which is how my book shows it)
From what I gather, if I am looking at this correctly:
Conditions 1 and 2 are not met, but condition 3 is
$f(2)$ both parts of the function $x^2=4$ and $3x-2=4$ equal 4 but only $3x-2$ meets the if $x>2$. So the $f(2)=4 \ne2$ 
The third condition is true because they are equal?
$lim_{x\to 2}(x^2)=4$
$lim_{x\to 2}(3x-2)=4$
So, I think I am on the right track, but not positive.
 A: Some books do not include the first criteria.
A function is continuous if $\lim_\limits{x\to a} f(x) = f(a)$ for all $a$ in the domain of $f$  
This also concludes that the limit exists at all $a$ in the domain.
If the limit exists then $\lim_\limits{x\to a^-} f(x) = \lim_\limits{x\to a^+} f(x)$
By this definition $f(x) = \frac 1x$ is continuous everywhere the function is defined.
But, using the definition you have provided.
$2$ is not in the domain if $f(x)$ and the function is not continuous (and you can stop here).
As for the limit.
$\lim_\limits{x\to a^-} = \lim_\limits{x\to a^+} = 4$  
The limit is defined.  Condition 2 is met.  Condition's 1 and 3 have not been met.
A: You are right that condition 1 is not met. Since continuity requires all three conditions to be met, you already have enough information to conclude that the function is discontinuous at $2$. 
On the other hand, condition 2 is met.  As for condition 3, it makes no sense unless conditions 1 and 2 are both met, since it talks about "the two values" obtained from those conditions. But nothing in the present paragraph alters the fact that the function is discontinuous at $2$ simply because it isn't defined at $2$.
