Graph of $\quad\frac{x^3-8}{x^2-4}$. I was using google graphs to find the graph of $$\frac{x^3-8}{x^2-4}$$ and it gave me:

Why is $x=2$ defined as $3$? I know that it is supposed to tend to 3. But where is the asymptote???

There is no asymptote at $x=2$. Note that 
$$\frac{x^3-8}{x^2-4}=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}.$$ 
For $x\ne 2$, we can cancel the $x-2$.
So near $x=2$ our function is very well behaved, it has a nice limit.  The singularity at $2$ is called a removable singularity. If we define a new function $g(x)$ by $g(x)$ equal to our given expression when $x\ne 2$, and $g(2)= 3$, the function $g(x)$ is very nice everywhere except at $x=-2$. (The singularity at $x=-2$ is not removable.)
Many pieces of graphing software completely ignore removable singularities. At least Alpha had the decency to put a dot there. 
