Conditions of Continuity (Limits) On a math test, for my online Honors Pre-Calculus course, that I recently took I got this question wrong and don't understand the explanation:

Suppose $f(x) = \begin{cases} x^2-2, & x \not= 2 \\ 0, & x = 2 \end{cases}$.

Which condition of continuity is not met by $f(x)$ at $x = 2$?
The correct answer is:Condition three: $\displaystyle \lim_{x \to c} f(x)=f(c)$
And the explanation is this:

  
*
  
*$f(2)=0$, so $f(c)$ is defined.
  
*$\displaystyle \lim_{x \to 2} f(x)=2$, so $\displaystyle \lim_{x \to 2} f(x)$ exists.
  
*$2 \not= 0$, so $\displaystyle \lim_{x \to c} f(x) \not= f(c)$.
  

Since the limit does not equal the function, the function is not continuous. The third condition was broken.

I don't understand this explanation because doesn't $c$ equal $2$ because limits are described as $x \to c$, in which case $f(x)=2$ and $f(c)=2$. Why does $f(c)=0$?
 A: If the function was continous,the condition:
$$\lim_{x \to 2} f(x)=f(2)$$
would be satisfied.
In this case, $\lim_{x \to 2} f(x)=2$ and $f(2)=0$
This is the reason why the function is not continuous.
If it was $f(2)=2$,it would be $\lim_{x \to 2} f(x)=f(2)$,and so the function would be continuous.
EDIT:
$x^2-1$ is continuous as a polynomial.
$0$ is continuous as a constant.
In order $f$ to be continuous,we have to check the continuity at $x=2$,where the value of the function changes.
In order the function to be continuous at $x=2$, it must be: $ \lim_{x \rightarrow 2^{-}}f(x)=\lim_{x \rightarrow 2^{+}} f(x)=f(2)$
$$\lim_{x \rightarrow 2^{-}}f(x)=\lim_{x \rightarrow 2^{-}} x^2-2=2$$
$$\lim_{x \rightarrow 2^{+}} f(x)=\lim_{x \rightarrow 2^{+}} x^2-2=2$$
$$f(2)=0$$
As $ \lim_{x \rightarrow 2^{-}}f(x)=\lim_{x \rightarrow 2^{+}} f(x) \neq f(2)$,the function is not continuous at $x=2$.
A: When we are looking at limits of functions we want to see what is value of function converge to as we approach $c$ (not at the point itself) thus for any sequence $\{a_{j}\}_{j=1}^{\infty}$ such that $a_{j}\rightarrow c$ (ie the sequence converges to c) we have that $\{f(a_{j})_{j=1}^{\infty}\}=\{0,0,0,..\}$ we have $f(a_{j})\rightarrow 0$ 
