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

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).

 $\lim_{x\rightarrow a}f(x)$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.

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

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.

 $\lim_{x\rightarrow a}f(x)$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 functionis 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.

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.

## Jump Discontinuity

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.

 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 valuesof 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) = $\left\{\begin{matrix} \frac{1}{2}x^{2} & if & x<2\\ x+1 & if & x\geq 2 \end{matrix}\right.$

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

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.

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

## Examples of Discontinuity

### 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.