Continuity and Discontinuity: Definition, Examples, Questions

Continuity and Discontinuity – Definition and Concept

In mathematics, the concepts of continuity and discontinuity help us understand how smoothly a function behaves. Suppose $f$ is a real function defined on a subset of real numbers, and let $c$ be a point in the domain of $f$. Then, $f$ is said to be continuous at $x = c$ if

$\lim_{x \rightarrow c} f(x) = f(c)$

In simple words, a function is continuous at a point if the value of the function and its limit at that point are the same.

Definition of Continuity at a Point

More elaborately, if the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at $x = c$ all exist and are equal, then the function $f(x)$ is continuous at $x = c$.

If both the right-hand and left-hand limits at $x = c$ are equal, the common value is called the limit of the function at $x = c$.

Hence, the definition of continuity can be rephrased as:

A function is continuous at $x = c$ if it is defined at $x = c$ and its value equals the limit of the function at that point.

That is, $\lim_{x \rightarrow c} f(x) = f(c)$

Definition of Discontinuity

If the above condition is not satisfied, i.e., the limit of the function does not equal its value at that point, then $f$ is discontinuous at $x = c$, and $c$ is called a point of discontinuity of the function.

Graphical Understanding of Continuity and Discontinuity

If a function is continuous, its graph is a smooth, unbroken curve — you can draw it without lifting your pen.
On the other hand, for discontinuous functions, the graph has a break or gap.

A real function is continuous at a fixed point if you can draw its graph around that point without lifting the pen from the paper. But if you have to lift the pen at a point (say $x = a$), it means the function has a discontinuity there.

Continuity at a Point and Over an Interval

Many functions have graphs that do not break anywhere; such functions are called continuous functions.
Some functions, however, have one or more points where the graph breaks. These functions may still be continuous over certain intervals within their domains and discontinuous at specific points.

Therefore, continuity can be classified in two ways:

• Continuity at a Point – When the function satisfies the condition of continuity at a single value of $x$.

• Continuity over an Interval – When the function remains continuous for every value of $x$ in a given interval.

Continuity at a point

Let us see different types of conditions to see continuity at point $x = a$

We see that the graph of $f(x)$ has a hole at $x=a$, which means that $f(a)$ is undefined. At the very least, for $f(x)$ to be continuous at $x=a$, we need the following conditions:

(i) $f(a)$ is defined

Next, for the graph given below, although $f(a)$ is defined, the function has a gap at $x=a$. In this graph, the gap exists because lim $\lim\limits _{x → a }f(x)$ does not exist. We must add another condition for continuity at $x=a$, which is

(ii) $\lim\limits _{x \rightarrow a} f(x)$ exists

The above two conditions by themselves do not guarantee continuity at a point. The function in the figure given below satisfies both of our first two conditions but is still not continuous at $a$. We must add a third condition to our list:

(iii) $\lim\limits _{x \rightarrow a} f(x)=f(a)$

So, a function $f(x)$ is continuous at a point $x = a$ if and only if the following three conditions are satisfied:

i) $f(a)$ is defined

ii) $\lim\limits_{x \rightarrow a} f(x)$ exists

iii) $\lim\limits_{x \rightarrow a} f(x)=f(a)$ or

$\begin{aligned} & \lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a) \\ & \text { i.e. } \text { L.H.L. }=\text { R.H.L. }=\text { value of the function at } x=a\end{aligned}$

A function is discontinuous at a point $a$ if it fails to be continuous at $a$.

Continuity over an Interval

Over an open interval $(a, b)$

A function $f(x)$ is continuous over an open interval $(a, b)$ if $f(x)$ is continuous at every point in the interval.

For any $c \in(a, b), f(x)$ is continuous if

$
\lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c^{+}} f(x)=f(c)
$
Over a closed interval $[a, b]$

A function $f(x)$ is continuous over a closed interval of the form $[a, b]$ if
- it is continuous at every point in $(a, b)$ and
- is right-continuous at $x=a$ and
- is left-continuous at $x=b$.
i.e.At $\mathrm{x}=\mathrm{a}$, we need to check $f(a)=\lim _{x \rightarrow a^{+}} f(x)\left(=\lim _{h \rightarrow 0^{+}} f(a+h)=\right.$ R.H.L. $)$.

L.H.L. should not be evaluated to check continuity of the first element of the interval, $x=a$

Similarly, at $\mathrm{x}=\mathrm{b}$, we need to check $f(b)=\lim _{x \rightarrow b^{-}} f(x)\left(=\lim _{h \rightarrow 0^{+}} f(b-h)=\right.$ L.H.L. $)$.

R.H.L. should not be evaluated to check continuity of the last element of the interval $x=b$

Consider one example,

$f(x)=[x]$, prove that this function is not continuous in $[2,3]$,
Sol.
Condition 1
For continuity in $(2,3)$
At any point $x=c$ lying in $(2,3)$,
$f(c)=[c]=2($ as $c$ lies in $(2,3))$
LHL at $\mathrm{x}=\mathrm{c}: x \rightarrow \mathrm{c}^{-}[x]=2$ (as in close left neighbourhood of $\mathrm{x}=\mathrm{c}$, the function equals 2)

RHL at $\mathrm{x}=\mathrm{c}: x \rightarrow \mathrm{c}^{+}[x]=2$ (as in close right neighbourhood of $\mathrm{x}=\mathrm{c}$, the function equals 2)

So function is continuous for any c lying in $(2,3)$. Hence the function is continuous in $(2,3)$
Condition 2
Right continuity at $x=2$

$
\begin{aligned}
& f(2)=2 \\
& \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}[x]=\lim _{h \rightarrow 0^{+}}[2+h]=2
\end{aligned}
$

So $f(x)$ is left continuous at $x=2$
Condition 3
Left continuity at $x=3$
$f(3)=3$ and

$
\lim _{x \rightarrow 3^{-}}[x]=\lim _{h \rightarrow 0^{+}}[3-h]=2
$

(as in left neghbourhood of $3, f(x)=2$ )
So $f(3)$ does not equal LHL at $x=3$
hence $f(x)$ is not left continuous at $x=3$

So the third condition is not satisfied and hence $f(x)$ is not continuous in $[2,3]$

Types of Discontinuity

A function can become discontinuous in different ways depending on how its value or limit behaves at a point. Below are the main types of discontinuities. Understanding these helps in graph interpretation and solving calculus questions in Class 12, JEE, and CUET exams.

1. Removable Discontinuity

A removable discontinuity occurs when the limit of a function exists at a point, but the function is either not defined there or defined with a different value.
In other words, the “hole” in the graph can be filled by redefining the function.

Mathematically, $f(x)$ has a removable discontinuity at $x = c$ if

$\lim_{x \rightarrow c} f(x)$ exists but $\lim_{x \rightarrow c} f(x) \ne f(c)$

Example:
Let $f(x) = \dfrac{(x^2 - 1)}{(x - 1)}$ for $x \ne 1$.
Here, $\lim_{x \rightarrow 1} f(x) = 2$ but $f(1)$ is not defined.
By defining $f(1) = 2$, the discontinuity can be removed.

2. Jump Discontinuity

A jump discontinuity happens when the left-hand limit and right-hand limit at a point both exist but are not equal.
This results in a sudden “jump” in the graph.

Mathematically, a function $f(x)$ has a jump discontinuity at $x = c$ if

$\lim_{x \rightarrow c^-} f(x) \ne \lim_{x \rightarrow c^+} f(x)$

Example:
Consider $f(x) = \begin{cases}
2, & x < 1 \
5, & x \ge 1
\end{cases}$

Here, $\lim_{x \rightarrow 1^-} f(x) = 2$ and $\lim_{x \rightarrow 1^+} f(x) = 5$.
Since both limits are not equal, the function has a jump discontinuity at $x = 1$.

3. Infinite (Essential) Discontinuity

An infinite discontinuity occurs when the function approaches infinity (or negative infinity) near a certain point.
In this case, the limit of the function does not exist in a finite sense.

Mathematically, a function $f(x)$ has an infinite discontinuity at $x = c$ if

$\lim_{x \rightarrow c^-} f(x) = \infty$ or $\lim_{x \rightarrow c^+} f(x) = \infty$

Example:
Let $f(x) = \dfrac{1}{x - 2}$.
As $x \rightarrow 2$, the denominator tends to zero, and $f(x)$ tends to infinity.
Hence, the function has an infinite discontinuity at $x = 2$.

Table of Types of Discontinuity

Solved Examples Based on Continuity and Discontinuity

Example 1: Let $a, b \in \mathbf{R}, (a \neq 0)$. If the function is defined as

$f(x)= \begin{cases}
\frac{2x^2}{a}, & 0 \leq x < 1 \\
a, & 1 \leq x < \sqrt{2} \\
\frac{2b^2 - 4b}{x^3}, & \sqrt{2} \leq x < \infty
\end{cases}$

and is continuous in the interval $[0, \infty)$, then the ordered pair $(a, b)$ is:

1. $(\sqrt{2}, 1-\sqrt{3})$

2. $(-\sqrt{2}, 1+\sqrt{3})$

3. $(\sqrt{2}, -1+\sqrt{3})$

4. $(-\sqrt{2}, 1-\sqrt{3})$

Solution:

Since the function is continuous,
it must be continuous at $x = 1$ and $x = \sqrt{2}$.

$\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^+} f(x) = f(1)$

$\Rightarrow \frac{2}{a} = a$

$\Rightarrow a^2 = 2$

and

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

$\Rightarrow a = \frac{2b^2 - 4b}{2\sqrt{2}}$

$\Rightarrow b^2 - 2b = \sqrt{2}a$

If $a = \sqrt{2}$, then $b^2 - 2b - 2 = 0 \Rightarrow b = 1 \pm \sqrt{3}$

If $a = -\sqrt{2}$, then $b^2 - 2b + 2 = 0$, which gives imaginary $b$.

Hence, the ordered pair $(a, b)$ is $(\sqrt{2}, 1+\sqrt{3})$ or $(\sqrt{2}, 1-\sqrt{3})$.

Example 2: If the function

$g(x) = \begin{cases}
k\sqrt{x+1}, & 0 \leq x \leq 3 \\
mx + 2, & 3 < x \leq 5
\end{cases}$

is differentiable, then the value of $k + m$ is:

1. $2$

2. $\frac{16}{5}$

3. $\frac{10}{3}$

4. $4$

Solution:

Condition for Continuity

A function $f(x)$ is continuous at $x = a$ if:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists (i.e., $LHL = RHL$).
3. $\lim_{x \to a} f(x) = f(a)$.

Condition for Differentiability

A function $f(x)$ is differentiable at $x = a$ if both $L f'(a)$ and $R f'(a)$ exist and are equal.

For continuity at $x = 3$,

$k \times 2 = 3m + 2$

For differentiability at $x = 3$,

$\frac{k}{2\sqrt{x+1}} \big|_{x=3} = m$

$\Rightarrow k = 4m$

Solving both,

$m = \frac{2}{5}$ and $k = \frac{8}{5}$

Therefore, $m + k = 2$.

Example 3: Consider the function $f(x) = [x] + [1 - x]$, $-1 \leq x \leq 3$,
where $[x]$ is the greatest integer function.

Statement I: $f$ is not continuous at $x = 0, 1, 2, 3$
Statement II:
$f(x) = \begin{cases}
-x, & -1 \leq x < 0 \\
1 - x, & 0 \leq x < 1 \\
1 + x, & 1 \leq x < 2 \\
2 + x, & 2 \leq x \leq 3
\end{cases}$

1. Statement I is true; Statement II is false.

2. Statement I is true; Statement II is true; but not a correct explanation.

3. Statement I is true; Statement II is true; and a correct explanation.

4. Statement I is false; Statement II is true.

Solution:

Let $f(x) = [x] + [1 - x]$, where $[x]$ is the greatest integer function.

$f$ is not continuous at $x = 0, 1, 2, 3$.
But in Statement II, $f(x)$ is continuous at $x = 3$.

Hence, Statement I is true and Statement II is false.

Answer: Option 1

Example 4: $f(x) = \begin{cases}
(x - 1)^{\frac{1}{2 - x}}, & x > 1, x \ne 2 \\
k, & x = 2
\end{cases}$

Find $k$ for which $f$ is continuous at $x = 2$.

1. $1$

2. $e$

3. $e^{-1}$

4. $e^{-2}$

Solution:

For a function to be continuous at $x = a$,

$L = R = V$

That is,

$LHL = RHL = f(a)$

For $x = 2^{+}$,

$\lim_{x \rightarrow 2} (x - 1)^{\frac{1}{2 - x}} = \lim_{x \rightarrow 2} (1 + (x - 2))^{\frac{1}{2 - x}}$

$= e^{\frac{x - 2}{2 - x}} = e^{-1}$

Thus, $k = e^{-1}$.

Answer: Option 3

Example 5: Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as

$f(x) = \begin{cases}
\frac{\sin((a+1)x) + \sin(2x)}{2x}, & x < 0 \\
b, & x = 0 \\
\frac{\sqrt{x + bx^3} - \sqrt{x}}{bx^{5/2}}, & x > 0
\end{cases}$

If $f(x)$ is continuous at $x = 0$, find $a + b$.

1. $-\frac{3}{2}$

2. $-3$

3. $-\frac{5}{2}$

4. $-2$

Solution:

Since $f(x)$ is continuous at $x = 0$,

$\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)$

$f(0) = b$

Now,

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left( \frac{\sin((a+1)x)}{2x} + \frac{\sin(2x)}{2x} \right)$

$= \frac{a + 1}{2} + 1$

and

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x + bx^3} - \sqrt{x}}{bx^{5/2}}$

$= \lim_{x \to 0^+} \frac{x + bx^3 - x}{bx^{5/2}(\sqrt{x + bx^3} + \sqrt{x})}$

$= \lim_{x \to 0^+} \frac{bx^3}{bx^{5/2}(\sqrt{x + bx^3} + \sqrt{x})}$

$= \frac{1}{2}$

From both limits,

$\frac{1}{2} = b = \frac{a + 1}{2} + 1$

$\Rightarrow b = \frac{1}{2}, ; a = -2$

Hence, $a + b = -\frac{3}{2}$.

Answer: Option 1

List of topics related to Continuity and Discontinuity

This section gives you a complete list of all the important subtopics connected to continuity and discontinuity in Class 12 Maths. You’ll find what each topic covers and how it links to differentiation, composite functions, and graphical interpretation - perfect for quick revision before exams.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Derivative as Rate Measure: Definition, Formula, Examples

Continuity of Composite Function

Continuity And Differentiability

Differentiability of Composite Function

NCERT Resources

This section brings together all NCERT-based study materials related to Continuity and Differentiability - including notes, solutions, and exemplar problems. It’s your one-stop spot for NCERT Class 12 Maths Chapter 5 preparation.

NCERT Class 12 Maths Notes for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Solutions for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Exemplar Solutions for Chapter 5 - Continuity and Differentiability

Practice Questions based on Continuity and Discontinuity

This section includes a curated list of practice questions and MCQs to test your understanding of continuity, discontinuity, and differentiability. These questions help reinforce theory with practical application.

Continuity And Discontinuity - Practice Question MCQ

We have shared below the links to practice questions on the related topics of continuity and discontinuity: