What Can We Say About the Continuity of $y=\frac{x}{x}$ at $x=0$? If we can't divide by $0$, should $\frac{x}{x}$ be discontinuous and undefined at $x=0$ or is it continuous with value $1$? Most online graph calculators plot a continuous curve. 

If it's continuous at $x=0$ with $y=1$, then we should be able to say that $\frac{(a-b)}{(a-b)} = 1$ at $a=b$. Or $q^2*\frac{p}{q}$ is $0$ at $q = 0$. And that whole proof of $2=1$ would hold true. The graphs online for say $\frac{x^2-4}{x-2}$ at $x=2$ are puzzling me.
Haha. That's what we call a removable discontinuity. 

Now $f(x)=1$ and $g(x)=\displaystyle \frac{x}{x}$ are very similar functions, with one difference you just pointed out. (Discontinuity at $x=0$)

A similar thing for $f(x) = x-2$ and $g(x) = \displaystyle \frac{x^2-4}{x+2}$. (Discontinuity at $x=-2$.)

We notice that even at $x=-1.9999$ the values of the function are the same. We say that the limit of $g(x)$ as $x$ approaches $-2$ exists, even if, as we can see, $g(-2)$ does not exist. 

A nonremovable discontinuity would be something like, $\displaystyle \frac{1}{x}$ as $x$ approaches $0$. 

I hope that helps you understand. It is still a discontinuity, but since it's removable, we can sometimes patch it up. 

Consider the piecewise function $h(x)$. 

$h(x)=\displaystyle \begin{cases} 
      \frac{x}{x} &amp; x \neq 0 \\
      1 &amp; x=0
   \end{cases}$

We were only able to "patch up" the discontinuity because it was removable. It was a infinitesimally small hole, not an entire asymptote like in the case of $f(x) = \displaystyle \frac{1}{x}$.
You can't divide by $0$, so the value at $0$ is undefined. The screens of graphing calculators have a finite resolution and although a hole should be in the graph, the hole might be "in between" two actual pixels and may not be visible. With high resolution graphs, even if it's displayed it may be too small to see. There are many reasons we don't rely on visuals to make deductions and this is one of them.