discontinuous function

In a removable discontinuity, \[\lim_{x\to a}f(x)\] exists, however \[\lim_{x\to a}f(x)\ne f(a)\]. This may be due to \[f(a)\] being undefined (first example), or \[f(a)\] having a "wrong value" (second example). The discontinuity can be eliminated by changing the definition of \[f(x)\] at \[a\], thus the name "removable".

In the left-most graph, if we modified \[f(x)\] to:

\begin{align*} f\left(x\right)= \begin{cases} \displaystyle\frac{x^{2}-1}{x-1} & x\ne1\\ 1 & x=1\\ \end{cases} \end{align*}

the function would now be a continuous one.

The right and left-hand limits exist and are finite, but are not equal. Since \[\lim_{x\to a^{-}}f(x)\ne\lim_{x\to a^{+}}f(x)\], the limit \[\lim_{x\to a}f(x)\] does not exist (as explained in here). Quite uncommon in simple formulas, however they appear frequently in fields such as engineering.

Both of the one-sided limits exist perhaps as \[\infty\] or \[-\infty\], and at least one of them must be \[\pm\infty\].

An essential discontinuity is one which isn't of the three previous types, at least one of the one-sided limits doesn't exist, in this case the function infinitely oscillates near 0. Neither \[\lim_{x\to 0^{-}}f(x)\] nor \[\lim_{x\to 0^{+}}f(x)\] exists, and \[\lim_{x\to 0}f(x)\] is undefined as \[f(0)\] is undefined. In applications, these kind of functions are rarely encountered as nature doesn't have much use for such functions.

Let \[f(x)\]:

\begin{align*} f(x)= \begin{cases} 1 & x>0\\ -1 & x<0\\ \end{cases} \end{align*}

Although we're inclined to assume that \[f^{\prime}(x)=0\] for the entire graph, this is not so as \[f^{\prime}(0)\] does not exist. By definition,

\begin{align*} f^{\prime}(0)=\lim_{\Delta x\to0}\frac{f(0+\Delta x)-f(0)}{\Delta x}\\ \end{align*}

but \[f(0)\] does not even exist. Therefore \[f^{\prime}(0)\] would be undefined.