Why is the function continuous at a point which gives the case 0/0? I have this function : $f(x) = \frac{6x^2+18x+12}{x^2-4}$, the domain is R. How come its graph is continuous at $x = -2$? I know it can be simplified to $\frac{6(x+1)}{x-2}$ ( firstly $f(x) = \frac{6(x+1)(x+2)}{(x+2)(x-2)}$ ). But for this simplification $x$ must not be $-2$ or $+2$. 
If I plot the function using different sites on the internet $f$ is continuous at $x = -2$. Why is that true? If $x = -2$ it is the $0/0$ case, even if for the simplification it's not. The simplification implies that $x$ is not $-2$.
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIoNip4XjIrMTh4KzEyKS8oeF4yLTQpIiwiY29sb3IiOiIjMDAwMDAwIn0seyJ0eXBlIjoxMDAwLCJ3aW5kb3ciOlsiLTI2IiwiMjYiLCItMTYiLCIxNiJdfV0-
 A: The function is not defined at the point $-2$, but it can be extended continuously at this point. As you have seen for $x \in \mathbb{R}-\{2,-2\}$ you have
$$f(x)=6\frac{(x+1)(x+2)}{(x+2)(x-2)}=6\frac{x+1}{x-2}$$
Now as $x\to-2$ the expression will always approach $6\frac{-1}{4}=-3/2$. This means that the function $g: \mathbb{R}-\{2\} \to \mathbb R$, $$g(x)=\begin{cases}f(x) & x \neq -2\\ -3/2 & x=-2\end{cases}$$
is continuous, and if you restrict it to the domain of $f$ you will get $f$.
$$g\lvert_{\mathbb R -\{2,-2\}} = f$$
This is then example of the concept of the continuous continuation of a function to a larger domain.
A: Here is a simpler example with the same features.
Define the function $g: (-\infty,0) \cup (0,\infty) \to \mathbb R$ by $$g(x)=\frac{x}{x}$$
 This is not continuous at $0$, because it is not even defined at $0$. But obviously it can be simplified to $g(x)=1$ on its domain, where $x$ is non-zero. So the function $h:\mathbb R \to \mathbb R$ defined by
$$h(x)=1$$
is an extension of $g$ to the whole of $\mathbb R$, that is continuous everywhere.
