What Can We Say About the Continuity of $y=\frac{x}{x}$ at $x=0$?

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 $h(x)$.

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

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 $f(x) = \displaystyle \frac{1}{x}$.