Types of Discontinuities

Discontinuity is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which graphs of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of Continuity and Discontinuity have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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In this article, we will cover the concepts of Non-removable, Infinite, and Oscillatory Type Discontinuity. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of one question has been asked on this concept in 2023.

Types of Discontinuity

A non-continuous function is said to be a discontinuous function.

There are various kinds of discontinuity at a point, which are classified as shown below:

1. Removable Discontinuity

• Missing point discontinuity
• Isolated point discontinuity

2. Non-Removable Discontinuity

• Finite Type

• Infinite Type

• Oscillatory

Removable Discontinuity

In this type of discontinuity, the limit of the function $\lim\limits_{x \rightarrow a} f(x)$ necessarily exists but it is either not equal to $\mathrm{f}(\mathrm{a})$ or $\mathrm{f}(\mathrm{a})$ is not defined. However, it is possible to redefine the function at $\mathrm{x}=\mathrm{a}$ in such a way that the limit of the function at $\mathrm{x}=\mathrm{a}$ is equal to $\mathrm{f}(\mathrm{a})$, i.e. $\lim\limits_{x \rightarrow a} f(x)=f(a)$.

Again removable discontinuity can be classified into missing point discontinuity and isolated point discontinuity

Missing point discontinuity: In this type of removable discontinuity, the limit of the function exist at a particular point but the function is not defined.

Isolated point discontinuity: In this type of removable discontinuity, the limit of the function exist at a particular point and the function is also defined but the limit and the function defined will not be equal.

Consider the function $f(x)=\frac{x^2-4}{x-2}$, where, $x \neq 2$ $\therefore \quad f(x)=\frac{(x+2)(x-2)}{(x-2)}=(x+2), x \neq 2$

In the above graph, observe that the graph has a hole (missing point) at $x$ $=2$, which makes it discontinuous at $x=2$.

Here $\mathrm{LHL}=\mathrm{RHL}=4$
Here function $f(x)$ is not defined at $x=2$, i.e. $f(2)$ is not defined. However, we can redefine the function as

$
f(x)= \begin{cases}\frac{x^2-4}{x-2}, & x \neq 2 \\ 4, & x=2\end{cases}
$
It makes the function continuous as now $L H L=R H L=f(2)$.
A function f is said to possess removable discontinuity if at $\mathrm{x}=\mathrm{a}$ :

$
L=R \neq V_{x \rightarrow a^{-}} f(x)=\lim\limits_{x \rightarrow a^{+}} f(x) \neq f(a)
$

Non - Removable Discontinuity

In this type of discontinuity, the limit of the function at $\mathrm{x}= a$ does not exists, i.e. $\lim\limits_{x \rightarrow a} f(x)$ does not exist, so it is not possible to redefine the function in any way to make it continuous at $\mathrm{x}=\mathrm{a}$.

Again non-removable discontinuity can be classified into three categories

Finite Discontinuity: In this type of discontinuity, both left-hand and right-hand limit exists but, they are unequal.
i.e. $\lim\limits_{x \rightarrow a^{-}} f(x)=L_1$ and $\lim\limits_{x \rightarrow a^{+}}=L_2$ but $L_1 \neq L_2$

For example

$
f(x)=\left\{\begin{array}{cc}
x^2, & x \leq 1 \\
x+1, & x>1
\end{array}\right.
$
Here, $\quad \lim\limits_{x \rightarrow 1^{+}} f(x)=\lim\limits_{x \rightarrow 1^{+}}(x+1)=2$

$
\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{x \rightarrow 1^{-}} x^2=1
$
At $x=1$, both LHL and RHL exist, but they are not equal

The value of the function jumps by " 1 " units at $x=1$, hence, such kinds of discontinuity are also known as jump discontinuity.

In this case, the non-negative difference between the two limits is called the jump of discontinuity. A function having a finite number of jumps in a given interval is called a piecewise continuous or sectionally continuous function

Example
$f(x) \tan ^{-1}\left(\frac{1}{x}\right)$ at $x=0$
$R H L$ i.e. $\left.f\left(0^{+}\right)=\frac{\pi}{2}\right\}$
$L H L$ i.e. $\left.f\left(0^{-}\right)=-\frac{\pi}{2} \quad\right\}$ jump $=\pi$

Finite non-removable discontinuity:

A function $f$ is said to possess finite irremovable discontinuity at $x=a$ if at $\mathrm{x}=\mathrm{a}$ the left hand limit both exist finitely but are unequal.

$
\lim\limits_{x \rightarrow a^{-}} f(x) \neq \lim\limits_{x \rightarrow a^{+}} f(x)
$

Infinite Discontinuity:

Here, at least one of the two limits (L.H.L. and R.H.L) is infinity or minus infinity

Consider the function,

$
f(x)=\frac{1}{x}
$
Here, L.H.L. $=\lim\limits_{x \rightarrow 0^{-}} \frac{1}{x}=-\infty$ and R.H.L. $=\lim\limits_{x \rightarrow 0^{+}} \frac{1}{x}=\infty$

The function has infinite type of discontinuity at $x = 0$

Oscillatory Type Discontinuity:

In this type of discontinuity, the limit of the function doesn't exist but oscillates between two quantities.

For example,

$
\begin{aligned}
& f(x)=\sin \left(\frac{1}{x}\right) \\
& \lim\limits_{x \rightarrow 0} f(x)=\text { a value between }-1 \text { to } 1
\end{aligned}
$

$\therefore$ limit doesn't exists as it's oscillates between -1 to 1 as $x \rightarrow 0$

Graph of $f(x)=\sin (1 / x)$

Infinite non-removable discontinuity

A function $f$ is said to possess infinite irremovable discontinuity at $x=a$ the left-hand limit and right-hand limit both do not exist and are infinite.

Recommended Video Based on Types of Discontinuity

Solved Examples Based On the Non-Removable, Infinite, and Oscillatory Type Discontinuity:

Example 1: $f(x)= \begin{cases}x+2 & x \leq 1 \\ \{3-x & x>1\end{cases}$, then at $x=1$
1) $f(x)$ is continuous
2) $f(x)$ has removable discontinuity
3) $f(x)$ has non-removable discontinuity
4) None of these

Solution:

Here

$
\begin{aligned}
& L H L=\lim\limits_{x \rightarrow 1^{-}} x+2=3 ; \\
& R H L=\lim\limits_{x \rightarrow 1^{+}} 3-x=2 ; \\
& f(1)=3 \\
& L H L \neq R H L
\end{aligned}
$
So limit doesn't exist, hence non-removable discontinuity at $\mathrm{x}=1$
Hence, the answer is the option 3.

Example 2: Let $f(x)=\{1 /|x| ; x \neq 0$

$
\{0 ; x=0 \text { then at } \mathrm{x}=0
$

1) $f(x)$ is continuous
2) $f(x)$ has removable discontinuity
3) $f(x)$ is continuous from right of $x=0$
4) $f(x)$ has infinite nonremovable discontinuity

Solution:

As we have learned

Infinite irremovable discontinuity -A function $f$ is said to possess discontinuity at $\mathrm{x}=$ a the left-hand limit and right-hand limit both do not exist and are infinite.

- wherein

$
\begin{aligned}
L H L & =\lim\limits_{x \rightarrow 0^{-}} 1 /|x|=\infty \\
R H L & =\lim\limits_{x \rightarrow 0^{+}} 1 /|x|=\infty
\end{aligned}
$
LHL and RHL both are infinite
Example 3: $f(x)=[x] \forall n \epsilon R$, then $\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=2$ because
1) L.H.L =R.H.L but different from $f(x)$
2) L.H.L are not defined
3) R.H.L are not defined
4) L.H.L, R.H.L and $f(x)$ are not equal all together

Solution: As we have learned
The rule for continuous -
A function is continuous at $x=a$ if and only if

$
L=R=V
$

L.H.L R.H.L value at $\mathrm{x}=\mathrm{a}$.
- wherein

Where

$
\begin{aligned}
& L=\lim\limits_{x \rightarrow a^{-}} f(x) \\
& R=\lim\limits_{x \rightarrow a^{+}} f(x) \\
& V_I=\lim\limits_{x \rightarrow a} f(x) \\
& \lim\limits_{x \rightarrow 2^{-}} f(x)=\lim\limits_{x \rightarrow 2^{-}}[x]=1=L H L \\
& \lim\limits_{x \rightarrow 2^{+}} f(x)=\lim\limits_{x \rightarrow 2^{+}}[x]=2=R H L \\
& f(2)=[2]=2
\end{aligned}
$

$\therefore L H L, R H L$ and $\mathrm{f}(2)$ are all not equal together so Hence, the answer is the option 4.

Example 4: If $\mathrm{f}(\mathrm{x})=\quad f(x)=\left\{\begin{array}{cl}\frac{\sin \{\cos x\}}{x-\pi / 2} & , x \neq \pi / 2 \\ 1 & , x=\pi / 2\end{array}\right.$, where \{\} represents the fractional part function, then
1) $\mathrm{f}(\mathrm{x})$ is continuous at $x=\pi / 2$
2) $\lim\limits_{x \rightarrow \pi / 2} \mathrm{f}(\mathrm{x})$ exists, but f is not continuous at $x=\pi / 2$
3) $\lim\limits_{x \rightarrow \pi / 2} f(x)$ does not exists
4) $\lim\limits_{x \rightarrow \pi / 2} f(x)=1$

Solution:

Continuity at a point -
A function $f(x)$ is said to be continuous at $x=a$ in its domain if
1. $f(a)$ is defined : at $x=a$.
2. $\lim\limits_{x \rightarrow a} f(x)$ exists means limit $x \rightarrow a$
of $f(x)$ at $x=$ a exists from left and right.
3. $\lim\limits_{x \rightarrow a} f(x)=f(a)$ then the limit equals the value at $x=a$

Continuity from Left -
The function $f(x)$ is said to be continuous from the left at $x=$ an if

$
\lim\limits_{x \rightarrow a^{-}} f(x)=f(a)
$
Continuity from Right -
The function $f(x)$ is said to be continuous from right at

$
\begin{aligned}
& x=a: \text { if } \lim\limits_{x \rightarrow a^{+}} f(x)=f(a) \\
& \quad \text { R.H.L }=\lim\limits_{h \rightarrow 0^{+}} f(\pi / 2+h)=\lim\limits_{h \rightarrow 0^{+}} \frac{\sin (1-\sin h)}{h} \rightarrow \infty
\end{aligned}
$

$
\begin{aligned}
& \text { L.H.L }=\lim\limits_{h \rightarrow 0^{-}} f(\pi / 2-h)=\lim\limits_{h \rightarrow 0^{-}} \frac{\sin (\sin h)}{-h} \\
& \lim\limits_{h \rightarrow 0^{-}}\left(\frac{\sin (\sin h)}{\sin h} \times \frac{\sin h}{-h}\right)=1 *-1=-1 \\
& \text { L.H.L } \neq \text { R.H.L }
\end{aligned}
$
Hence, the answer is the option (3).

Example 5: If

$
f(x)=\left\{\begin{array}{cl}
\frac{e^{e / x}-e^{-c / x}}{e^{1 / x}+e^{-1 / x}} & x \neq 0 \\
k & x=0
\end{array}\right.
$

1) f is continuous at $x=0$ when $k=0$.
2) f is not continuous at $x=0$ for any real k .
3) $\lim\limits_{x \rightarrow 0} f(x)$ exist infinitely.
4) None of these

Solution:

$
\begin{aligned}
& \lim\limits_{x \rightarrow 0^{+}} \frac{e^{e / x}-e^{-e / x}}{e^{1 / x}+e^{-1 / x}}=\lim\limits_{x \rightarrow 0^{+}} \frac{e^{\frac{\epsilon-1}{\epsilon x}}\left(1-e^{-2 e / x}\right)}{\left(1+e^{-2 / x}\right)}=+\infty \\
& \lim\limits_{x \rightarrow 0^{-}} \frac{e^{e / x}-e^{-e / x}}{e^{1 / x}+e^{-1 / x}}=\lim\limits_{x \rightarrow 0^{-}} \frac{e^{-e / x}\left(e^{2 e / x}-1\right)}{e^{-1 / x}\left(e^{+2 / x}+1\right)} \\
& \lim\limits_{x \rightarrow 0^{-}} e^{-\left(\frac{c-1}{x}\right)}\left(\frac{e^{2 e / x}-1}{e^{2 / x}+1}\right)=-\infty
\end{aligned}
$
The limit doesn't exist, so the $f(x)$ is discontinuous.
Hence, the answer is the option 2.

Summary

Discontinuity is an important concept in Calculus. It provides a deep understanding of how the functions interact and change. It is very helpful in practical applications for physics, economics, etc. There are various kinds of discontinuity at a point. Overall, this provides a better interpretation of the functions leading to more accurate solutions.