What's going on at f(0)? Take the following simple function: $$f(x) = \frac{x^2}{x}$$
How should one interpret $f(0)$? Is it undefined? My limited (heh) understanding of the delta-epsilon definition suggests that it is both continuous and differentiable, yet undefined. Huh?
Is there some sort of order of operations rule that allows the fraction to be simplified before plugging 0 in for x in the denominator? That seems arbitrary. Where does one find that rule laid out?
It seems obvious that at $x=0$ it's undefined, but it also seems that we could integrate this function. 
 A: In its current state, the function is not defined at zero because $0/0$ is not defined. However, if we extend the definition via
$$f(x) = \left\{\begin{array}{lr} \frac{x^2}{x} & x \ne 0 \\ 0 & x = 0\end{array}\right.$$
then it's fairly easy to see that this is continuous and differentiable on the entirety of $\mathbb{R}$. Hence the discontinuity at zero is called removable, since we can remove it via adding a definition at the point of interest.
A: It is a removable singularity.  Your $f(x)$ is not defined at $x=0$, but $\lim_{x \to 0}f(x)=0$  It is reasonable to define a new function $g(x)=\begin {cases} f(x) & x \neq 0\\0&x=0 \end {cases}$ and note that $g(x)=x$
A: You can't speak of the function $f(x)=\frac{x^2}{x}$ as being either continuous or differentiable at $0$ since both of those notions require the function to be defined at $0$, which is not the case here.  What is defined is the limit $\lim_{x\to 0} f(x)$, which equals $0$. This is because the definition of the limit at a given point does not require the function to be defined at the point.
