Continuity And Differentiability: Definition, Formula, Examples

Continuity and Differentiability – Definition, Concepts, Formulas, and Examples

In everyday life, imagine you’re drawing a line without lifting your pencil from the paper - the curve flows smoothly from one point to another. That’s what continuity in mathematics feels like. Now imagine checking how steep that line is at every point - that’s differentiability. These two ideas together form the heart of calculus, helping us understand how functions behave and change in the real world.

What is Continuity?

Continuity tells us whether a function has any sudden jumps, gaps, or breaks. In simple terms, a function is continuous at a point if you can draw its graph around that point without lifting your pen.

Mathematically, a function $f(x)$ is continuous at $x = c$ if:

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

This means three things must hold:

1. $f(c)$ is defined.

2. $\lim_{x \to c} f(x)$ exists.

3. The value of the limit equals the function value at that point.

If any of these fail, the function is discontinuous at $x = c$.

Types of Continuity

This section explains the different kinds of continuity in functions — from continuity at a single point to continuity across an entire interval. You’ll learn how to identify whether a function is smooth, has jumps, or behaves differently at its boundaries.

Continuity at a Point

A function $f(x)$ is continuous at $x = a$ if both the left-hand limit (LHL) and right-hand limit (RHL) exist and are equal to the function’s value at that point.

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

If the left and right limits differ, the graph has a jump discontinuity. If the limit exists but does not equal $f(a)$, it’s a removable discontinuity.

Example:
Let $f(x) = \frac{x^2 - 4}{x - 2}$.
For $x = 2$, the function is not defined, but $\lim_{x \to 2} f(x) = 4$.
If we redefine $f(2) = 4$, the function becomes continuous.

Continuity Over an Interval

When we check continuity across a range of values, we talk about continuity over an interval.

• Open Interval $(a, b)$: The function is continuous at every point between $a$ and $b$.

• Closed Interval $[a, b]$: The function is continuous on $(a, b)$, right-continuous at $x = a$, and left-continuous at $x = b$.

Example:
For $f(x) = [x]$, the greatest integer function, the function is continuous in each open interval $(n, n+1)$ but discontinuous at integer points.

Properties of Continuous Functions

Continuous functions behave predictably under basic operations.

1. The sum, difference, and product of continuous functions are continuous.
   If $f(x)$ and $g(x)$ are continuous at $x = a$, then $f(x) \pm g(x)$ and $f(x)g(x)$ are also continuous at $x = a$.

2. The quotient $\dfrac{f(x)}{g(x)}$ is continuous at $x = a$ if $g(a) \neq 0$.

3. The composition $f(g(x))$ is continuous at $x = a$ if $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $g(a)$.

4. Continuous functions defined on closed intervals $[a, b]$ are bounded and achieve both maximum and minimum values (Extreme Value Theorem).

What is Differentiability?

Once a function is continuous, the next question is — can we find its rate of change at each point? That’s where differentiability comes in.

A function $f(x)$ is differentiable at $x = a$ if the derivative $f'(a)$ exists. It is defined as:

$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

If this limit exists and is finite, $f(x)$ has a well-defined tangent at that point. If the limit doesn’t exist or gives different values from the left and right, the function is not differentiable there.

Relationship Between Continuity and Differentiability

Every differentiable function is continuous, but not every continuous function is differentiable.

• If $f(x)$ is differentiable at $x = a$, then $f(x)$ is continuous at $x = a$.

• However, continuity alone doesn’t ensure differentiability.

Example:
$f(x) = |x|$ is continuous everywhere, but at $x = 0$,
Left derivative $= -1$ and Right derivative $= 1$,
so $f'(0)$ does not exist — it’s not differentiable at $x = 0$.

Rules of Differentiation

Differentiation follows a few basic rules that make calculations easier:

1. Sum Rule: $(f + g)' = f' + g'$

2. Difference Rule: $(f - g)' = f' - g'$

3. Product Rule: $(fg)' = f'g + fg'$

4. Quotient Rule: $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$, where $g(x) \neq 0$

5. Constant Rule: $(k \cdot f)' = k \cdot f'$ for any constant $k$

These form the foundation for more complex differentiation techniques like the chain rule and implicit differentiation.

Differentiability of Standard Functions

Here are some standard derivatives to remember:

• $(x^n)' = n x^{n-1}$

• $(\sin x)' = \cos x$

• $(\cos x)' = -\sin x$

• $(e^x)' = e^x$

• $(\log x)' = \dfrac{1}{x}$ for $x > 0$

• $(a^x)' = a^x \log a$

These results are used extensively in calculus problems, especially when verifying differentiability or finding tangent slopes.

Types of discontinuity

A discontinuity occurs when a function suddenly breaks, jumps, or becomes undefined at certain points. Depending on the way the function behaves near those points, discontinuities can be classified into three main types:

1. Removable Discontinuity:
   This happens when the limit of the function exists at a point, but the function’s actual value is either missing or different. On a graph, it looks like a small hole that can be “fixed” by redefining the function at that point.
   Example: $f(x) = \dfrac{x^2 - 1}{x - 1}$ is undefined at $x = 1$, but the limit as $x \to 1$ exists, making it a removable discontinuity.

2. Jump Discontinuity:
   A jump discontinuity appears when the left-hand limit (LHL) and right-hand limit (RHL) exist but are not equal. The graph literally “jumps” from one value to another.
   Example: The greatest integer function $f(x) = [x]$ shows jump discontinuities at every integer point.

3. Infinite Discontinuity:
   This occurs when the function approaches infinity (or negative infinity) near a certain point. The graph shoots up or down vertically, creating a vertical asymptote.
   Example: $f(x) = \dfrac{1}{x}$ has an infinite discontinuity at $x = 0$, where the function becomes unbounded.

Points of Non-Differentiability

A function may fail to be differentiable due to:

1. Corners or Cusps – sharp turns (e.g., $f(x) = |x|$ at $x = 0$)

2. Discontinuities – jumps or breaks in the graph

3. Vertical Tangents – where the slope becomes infinite

Understanding these points helps in analysing real-world curves that change direction abruptly.

Examples of Continuity and Differentiability

• Example 1: $f(x) = x^2$
  Continuous and differentiable everywhere since the graph is smooth.

• Example 2: $f(x) = |x|$
  Continuous for all $x$, but not differentiable at $x = 0$.

• Example 3: $f(x) = [x]$
  Discontinuous at all integer points due to jumps in its graph.

These examples are commonly seen in board exam questions and entrance tests.

Applications of Continuity and Differentiability

• Physics: To find velocity and acceleration (rate of change of displacement).

• Economics: To study cost, revenue, and profit functions.

• Engineering: To analyse stress-strain curves or optimise design functions.

• Biology and Medicine: To model population growth or dosage rates that vary smoothly over time.

Continuity And Differentiability in Mathematics: Solved Previous Year Questions

Question 1:

If the function

$
f(x)=\left\{\begin{array}{cl}
k_1(x-\pi)^2-1 & x \leq \pi \\
k_2 \cos x, & x>\pi
\end{array}\right. \text { is twice }
$

differentiable, then the ordered pair $\left(k_1, k_2\right)$ is equal to :

Solution:

$f(x)$ is continuous and differentiable $f\left(\pi^{-}\right)=f(\pi)=f\left(\pi^{+}\right)$

$
\begin{aligned}
& k_1((\pi)-\pi)^2-1=k_2 \cos (\pi) \\
& -1=k_2 \\
& f(x)=\left\{\begin{array}{l}
2 k_1(x-\pi) ; x \leq \pi \\
-k_2 \sin x ; x>\pi
\end{array}\right. \\
& \mathrm{f}^{\prime}\left(\pi^{-}\right)=\mathrm{f}^{\prime}\left(\pi^{+}\right) \\
& 0=0
\end{aligned}
$

so, differentiable at $x=0$

$
\begin{aligned}
& f^{\prime}(x)=\left\{\begin{array}{c}
2 k_1 \\
-k_2 \cos x ; x>\pi
\end{array}\right. \\
& f^{\prime}\left(\pi^{-}\right)=f^{\prime \prime}\left(\pi^{+}\right) \\
& 2 k_1=k_2 \\
& k_1=\frac{k_2}{2} \\
& \Rightarrow\left(k_1, k_2\right)=\left(\frac{1}{2}, 1\right)
\end{aligned}
$

Hence, the answer is $\left ( \frac{1}{2},1 \right )$.

Question 2:

Let $f: S \rightarrow S$ where $S=(0, \infty)$ be twice differnetiable function such that $f(x+1)=x f(x)$. If $g: S \rightarrow R$ be defined as $g(x)=\log _e f(x)$ then the value of $\left|g^{\prime \prime}(5)-g^{\prime \prime}(1)\right|$ is equal to:

Solution:

$
\begin{aligned}
& \ln f(x+1)=\ln (x f(x)) \\
& \ln f(x+1)=\ln x+\ln f(x) \\
& \Rightarrow \quad g(x+1)=\ln x+g(x) \\
& \Rightarrow \quad g(x+1)-g(x)=\ln x \\
& \Rightarrow \quad g^{\prime \prime}(x+1)-g^{\prime \prime}(x)=-\frac{1}{x^2}
\end{aligned}
$
Put $x=1,2,3,4$
$g^{\prime \prime}(2)-g^{\prime \prime}(1)=-\frac{1}{1^2}$
$g^{\prime \prime}(3)-g^{\prime \prime}(2)=-\frac{1}{2^2}$
$g^{\prime \prime}(4)-g^{\prime \prime}(3)=-\frac{1}{3^2}$
$g^{\prime \prime}(5)-g^{\prime \prime}(4)=-\frac{1}{4^2}$
Add all the equations we get

$
\begin{aligned}
g^{\prime \prime}(5)-g^{\prime \prime}(1) & =-\frac{1}{1^2}-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2} \\
\mid g^{\prime \prime}(5)-g^{\prime \prime}(1) & =\frac{205}{144}
\end{aligned}
$

Hence, the answer is the $\frac{205}{144}$.

Question 3:

If $x=t^2, y=t^3$, then $\frac{d^2 y}{d x^2}$ is:

Solution:

Given that, $x=t^2, y=t^3$

$
\begin{aligned}
& \Rightarrow \frac{d x}{d t}=2 t \text { and } \frac{d y}{d t}=3 t^2 \\
& \frac{d y}{d x}=\frac{d y / d t}{d x / d t} \\
& \frac{d y}{d x}=\frac{3 t^2}{2 t}=\frac{3 t}{2} \\
& \frac{d^2 y}{d x^2}=\frac{3}{2} \frac{d t}{d x} \\
& \frac{d^2 y}{d x^2}=\frac{3}{2} \cdot \frac{1}{2 t}\left(\because \frac{d x}{d t}=2 t \Rightarrow \frac{d t}{d x}=\frac{1}{2 t}\right) \\
& \frac{d^2 y}{d x^2}=\frac{3}{4 t}
\end{aligned}
$

Hence, the answer is the $\frac{3}{4t}$.

Question 4:

$y=\log \left(\frac{1-x^2}{1+x^2}\right)$ then $\frac{d y}{d x}$ is equal to:

Solution:

$\begin{aligned} & y=\log \frac{1-x^2}{1+x^2} \\ & \Rightarrow \mathrm{y}=\log \left(1-\mathrm{x}^2\right)-\log \left(1+\mathrm{x}^2\right) \\ & \Rightarrow \frac{d y}{d x}=\frac{1}{1-x^2} \frac{d}{d x}\left(1-x^2\right)-\frac{1}{1+x^2} \frac{d}{d x}\left(1+x^2\right) \\ & \Rightarrow \frac{d y}{d x}=\frac{1}{1-x^2} \cdot(-2 x)-\frac{1}{1+x^2} \cdot(2 x) \\ & \Rightarrow \frac{d y}{d x}=\frac{-2 x}{1-x^2}-\frac{2 x}{1+x^2} \\ & \frac{d y}{d x}=\frac{-2 x\left(1+x^2\right)-2 x\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)} \\ & \frac{d y}{d x}=\frac{-2 x-2 x^3-2 x+2 x^3}{\left(1-x^4\right)} \\ & \frac{d y}{d x}=\frac{-4 x}{\left(1-x^4\right)}\end{aligned}$

Hence, the answer is $\frac{-4 x}{1-x^{4}}$.

Question 5:

The function $f(x) = \frac{4-x^2}{4x-x^3}$ is:

Solution:

We have, $f(x)=\frac{4-x^2}{4 x-x^3}$
Now, $f(x)$ is discontinuous where $4 x-x^3=0$.

$
\Rightarrow x\left(4 x-x^2\right)=0 \Rightarrow x=0 \text { or } 4 x-x^3=0 \Rightarrow x=0 \text { or } x= \pm 2
$

$\Rightarrow f(x)$ is discontinuous at exactly three points.

Hence, the answer is "discontinuous at exactly three points".

Continuity and Differentiability in Different Exams

The chapter on Continuity and Differentiability is a key part of Class 12 Mathematics and builds upon the concepts of limits and derivatives. It is frequently tested in board exams and competitive examinations. The table below highlights the exam-wise focus areas, commonly asked topics, and preparation strategies.

List of topics related to Continuity and Differentiability

This section gives you a complete list of key subtopics and concepts covered under Continuity and Differentiability, helping you understand what areas to focus on for exams and practice.

• Differentiability and Existence of Derivative
• Examining Differentiability Using the Graph of a Function
• Derivative as Rate Measure: Definition, Formula, Examples
• Differentiability of Composite Function
• Rolle’s Theorem: Definition, Formula, Examples, Questions
• Derivative of A Function In Parametric Form
• Differentiation Rules: Definition, Formula, Examples
• Monotonicity And Extrema of Functions

NCERT Resources for Continuity And Differentiability

This section compiles all NCERT-based resources for Chapter 5 - Continuity and Differentiability, including structured notes, solved examples, textbook solutions, and exemplar questions for complete revision.

• 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

NCERT Subject-wise Resources

Get subject-wise NCERT study resources with detailed chapter-wise notes and ample practice material to support efficient and focused exam preparation from the following links

Important Books and Resources for Continuity and Differentiability

Check below the recommended books and study materials that cover Continuity and Differentiability in detail, helping students build strong conceptual clarity and practise a wide range of problems for board exams and competitive tests like JEE.

Practice Questions based on Continuity and Differentiability

Here you’ll find curated practice questions and topic-wise MCQs to test your understanding of continuity, differentiability, and related theorems.

Conclusion

Continuity and Differentiability form a crucial part of Class 12 Mathematics and are essential for understanding the behaviour of functions. A clear grasp of definitions, conditions, and examples helps in analysing smoothness and change accurately. With regular practice and proper revision, students can confidently solve problems related to this chapter in board and competitive examinations.