In real life discontinuity is experienced in the form of breaks and abrupt changes. Mathematically a function is said to be discontinuous when the graph of the function contains breaks and jumps. A function is said to be discontinuous at x = a if either of the following is true.

  1. f(a) is not defined.
  2. $\lim_{x\rightarrow a}f(x)$ does not exist.
  3. $\lim_{x\rightarrow a}f(x)$ $\neq$ f(a)

Discontinuity at a point is broadly classified into two types, removable and irremovable discontinuities.

Removable discontinuity is also known as point discontinuity. This occurs, when $\lim_{x\rightarrow a}f(x)$ exists and either f(a) is undefined or $\lim_{x\rightarrow a}f(x)$ $\neq$ f(a).

 Removable Discontinuity             
Removable Discontinuity Introduction
$\lim_{x\rightarrow a}f(x)$
exists, But f(a) is not defined.
The graph has a hole at x = a.
       
$\lim_{x\rightarrow a}f(x)$
exists and f(a) is defined. But
$\lim_{x\rightarrow a}f(x)$ $\neq$ f(a).
The graph has a hole at x = a.

Such discontinuities can be removed by defining or refining the function at the point of discontinuity.

Example:

Consider the rational function f(x) = $\frac{x^{2}-5x+6}{x-2}$

The domain of the function is (-$\infty$, 2) U (2, $\infty$)  and f(2) is undefined.
The $\lim_{x\rightarrow 2}f(x)$ can be found by canceling the zero rendering common factor x - 2.
The limit x ->2 exists but f(2) is undefined.
The graph of the function also shows a straight line with a hole at the point corresponding to x = 2.

         Discontinuity Refining Function

The discontinuity can be removed by redefining the function as follows:
$\left\{\begin{matrix}
\frac{x^{2}-5x+6}{x-2} & if & x\neq 2\\
 -1& if & x=2
\end{matrix}\right.$
Non removable discontinuity occurs at a point x = a when $\lim_{x\rightarrow a}f(x)$ does not exist. The function may or may not be defined at the point.

  Non Removable Discontinuity     
  What is Non Removable Discontinuity
$\lim_{x\rightarrow a}f(x)$
does not exist. But f(a)
is defined. A 'Jump' is seen
in the graph.
  $\lim_{x\rightarrow a}f(x)$
does not exist and the function
is also not defined at x = a.
The graph blows up at x = a.

The discontinuity caused by the non existence of the limit at a point cannot be removed by redefining the function.

Examples:
The graph of tangent function f(x) = tan(x) has non removable discontinuities when x is equal to an odd multiple of $\frac{\pi }{2}$.

The graph seems to be making an infinite jump disappearing on top and appearing at the bottom.

Non Removable Discontinuity Example
 
Another type of discontinuity is oscillating discontinuity. f(x) = $cos($$\frac{2\pi }{x}$$)$ oscillates rapidly at x = 0 to have a limit as x → 0.

Non Removable Discontinuity Problem
Jump discontinuity is a non removable discontinuity that occurs when a function f(x) is defined at x = a, but $\lim_{x\rightarrow a}f(x)$ does not exist.

This phenomena can be observed in the floor function f(x) = $\left \lfloor x \right \rfloor$ (Greatest integer function) and ceiling function
f(x) = $\left \lceil x \right \rceil$ (Least integer function) at every integer value of x.

 Jump Discontinuity    Jump Discontinuity Example
The right limit does not exist at integer values
of x for the floor function, which causes the jump discontinuities.
           
The left limit does not exist at integer values
of x for the ceiling function, which causes the
jump discontinuities.

Jump discontinuity also occurs in some piecewise defined functions.
Consider the piecewise defined function,
f(x) = $\left\{\begin{matrix}
\frac{1}{2}x^{2} & if & x<2\\
x+1 & if & x\geq 2
\end{matrix}\right.$

Piecewise Discontinuity

The function is defined at x = 2 on the straight line y = x + 1 i.e. f(2) = 3.

$\lim_{x\rightarrow 2^{-}}$ = $\frac{1}{2}$ $\times$ 4 = 2 and $\lim_{x\rightarrow 2^{+}}$ = 3

Since the left and right limits are not equal, the limit does not exist for the function.

This has caused the jump discontinuity.
Infinite discontinuity occurs to the graph of a function f(x) at a point x = a, when both the $\lim_{x\rightarrow a}f(x)$ does not exist and the function is not defined at x = a.

We saw that the graph of the tangent function has infinite discontinuities at odd multiples of $\frac{\pi }{2}$. The graph seems to disappear at the top and appears at the bottom around these points. The function has vertical asymptotes at the points of infinite discontinuity.

Infinite discontinuity is a common feature observed in the graphs of rational functions.

Suppose f(x) = $\frac{x}{(x-2)(x+3)}$
Infinite discontinuities occur at x = -3 and x = 2 where f(x) is not defined and also the limits do not exist.

   Infinite Discontinuity

The vertical asymptotes at the points of infinite discontinuities are shown in dotted lines.

Solved Examples

Question 1: Examine the discontinuity for f(x) = $\left\{\begin{matrix}
x^{2} & if & x<0\\
 sin(2x)& if & 0\leq x< 2\pi\\ 
x-2\pi  & if & x> 2\pi
\end{matrix}\right.$

If the discontinuity is removable, refine the function to make the function continuous.
Solution:
 
Let us check the continuity at two critical points x = 0 and x = 2$\pi$ around which the function definition changes.

$\lim_{x\rightarrow 0^{-}}f(x)$ = $\lim_{x\rightarrow 0}x^{2}$ = 0

$\lim_{x\rightarrow 0^{+}}f(x)$ = $\lim_{x\rightarrow 0}sin(2x)$ = 0

Thus $\lim_{x\rightarrow 0}f(x)$ = 0 and f(0) = sin(0) = 0 ⇒ $\lim_{x\rightarrow 0}f(x)$ = f(0) = 0

Hence the function is continuous at x = 0

$\lim_{x\rightarrow 2\pi ^{-}}f(x)$ = $\lim_{x\rightarrow 2\pi }sin(2x)$ = 0

$\lim_{x\rightarrow 2\pi ^{+}}f(x)$ = $\lim_{x\rightarrow 2\pi }(x-2\pi )$ = 2$\pi$ - 2$\pi$ = 0

Hence $\lim_{x\rightarrow 2\pi }f(x)$ exists  and equal to 0. But the function is not defined at x = 2$\pi$.
Hence the function has a hole at x = 2$\pi$. The discontinuity can be removed by replacing the inequality x > 2$\pi$ in the last function definition by x $\geq$ 2$\pi$. 

Thus the function is redefined with the point discontinuity removed as follows:

f(x) = $\left\{\begin{matrix}
x^{2} & if & x<0\\
 sin(2x)& if & 0\leq x< 2\pi \\
x-2\pi  & if & x\geq  2\pi
\end{matrix}\right.$
 

Question 2: Explain why f(x) = ln|x - 5| is discontinuous at x = 5
Solution:
 
f(5) = ln|5 -5| = ln 0  (undefined)

   $\lim_{x\rightarrow 5}ln|x-5|$ also does not exist.

   Hence the function has an infinite discontinuity at x = 5.