The Hook: Can You Draw Without Lifting Your Pen?
In Jawan (2023), Shah Rukh Khan’s team needs to cross a bridge. But what if there’s a gap in the middle?
You can walk on either side, but you can’t cross the gap without jumping.
This is exactly what discontinuity means in mathematics! A continuous function is like an unbroken bridge — you can trace it without lifting your pen. A discontinuous function has gaps, jumps, or breaks.
Can you trace $f(x) = \frac{1}{x}$ through $x = 0$ without lifting your pen? No! It has a discontinuity at $x = 0$.
Interactive: Explore Continuous Functions
Try plotting these and observe continuity:
- $f(x) = x^2$ (continuous everywhere)
- $f(x) = \frac{1}{x}$ (discontinuous at $x = 0$)
- $f(x) = \lfloor x \rfloor$ (jump discontinuities at integers)
The Core Concept
What is Continuity?
A function $f(x)$ is continuous at $x = a$ if:
$$\boxed{\lim_{x \to a} f(x) = f(a)}$$In simple terms: The function value at $a$ equals what you’d expect from approaching $a$.
The Three Commandments of Continuity
For $f(x)$ to be continuous at $x = a$, ALL THREE must be true:
- $f(a)$ exists — The function is defined at $a$
- $\lim_{x \to a} f(x)$ exists — LHL = RHL
- $\lim_{x \to a} f(x) = f(a)$ — The limit equals the function value
Break any one of these, and you have a discontinuity!
Defined at the point ✓
Exists as a limit ✓
Limit equals function value ✓
All three → Continuous!
Visualizing Continuity
Continuous Function Example
$$f(x) = x^2 \text{ at } x = 2$$- $f(2) = 4$ ✓ (defined)
- $\lim_{x \to 2} f(x) = 4$ ✓ (limit exists)
- $\lim_{x \to 2} f(x) = f(2)$ ✓ (limit equals value)
Continuous at $x = 2$!
Discontinuous Function Example
$$f(x) = \frac{x^2 - 4}{x - 2} \text{ at } x = 2$$- $f(2)$ = undefined ✗ (division by zero)
- $\lim_{x \to 2} f(x) = 4$ ✓ (limit exists)
- Can’t check since $f(2)$ doesn’t exist ✗
Discontinuous at $x = 2$!
Types of Discontinuity
1. Removable Discontinuity
Definition: $\lim_{x \to a} f(x)$ exists, but either:
- $f(a)$ is not defined, OR
- $f(a) \neq \lim_{x \to a} f(x)$
Why “removable”? We can “fix” it by redefining $f(a) = \lim_{x \to a} f(x)$.
Example:
$$f(x) = \frac{x^2 - 9}{x - 3}$$At $x = 3$:
- $f(3)$ is undefined
- $\lim_{x \to 3} f(x) = 6$ (factoring gives $x + 3$)
Fix: Define $f(3) = 6$, and the function becomes continuous!
2. Jump Discontinuity
Definition: LHL ≠ RHL at $x = a$.
The function “jumps” from one value to another.
Example: Greatest Integer Function
$$f(x) = \lfloor x \rfloor \text{ at } x = 2$$- LHL: $\lim_{x \to 2^-} \lfloor x \rfloor = 1$ (approaching from 1.9, 1.99, etc.)
- RHL: $\lim_{x \to 2^+} \lfloor x \rfloor = 2$ (approaching from 2.1, 2.01, etc.)
Since LHL ≠ RHL, there’s a jump at $x = 2$.
Example: Piecewise Function
$$f(x) = \begin{cases} x + 1 & \text{if } x < 2 \\ x + 5 & \text{if } x \geq 2 \end{cases}$$At $x = 2$:
- LHL = 3, RHL = 7
- Jump discontinuity!
3. Infinite Discontinuity
Definition: $\lim_{x \to a} f(x) = \pm \infty$
The function shoots to infinity at $x = a$.
Example:
$$f(x) = \frac{1}{x - 2} \text{ at } x = 2$$- $\lim_{x \to 2^-} f(x) = -\infty$
- $\lim_{x \to 2^+} f(x) = +\infty$
Vertical asymptote at $x = 2$.
4. Oscillatory Discontinuity
Definition: The function oscillates infinitely as $x \to a$, so the limit doesn’t exist.
Example:
$$f(x) = \sin\left(\frac{1}{x}\right) \text{ at } x = 0$$As $x \to 0$, $\frac{1}{x} \to \infty$, causing $\sin$ to oscillate infinitely between -1 and +1.
No limit exists!
Summary Table: Types of Discontinuity
| Type | Condition | Can We Fix It? | Example |
|---|---|---|---|
| Removable | Limit exists but ≠ $f(a)$ | Yes (redefine $f(a)$) | $\frac{x^2-1}{x-1}$ at $x=1$ |
| Jump | LHL ≠ RHL | No | $\lfloor x \rfloor$ at integers |
| Infinite | Limit = $\pm\infty$ | No | $\frac{1}{x}$ at $x=0$ |
| Oscillatory | Limit doesn’t exist (oscillation) | No | $\sin(1/x)$ at $x=0$ |
Continuity on an Interval
Continuity in an Interval
$f(x)$ is continuous on $(a, b)$ if it’s continuous at every point in $(a, b)$.
$f(x)$ is continuous on $[a, b]$ if:
- Continuous on $(a, b)$
- Right-continuous at $a$: $\lim_{x \to a^+} f(x) = f(a)$
- Left-continuous at $b$: $\lim_{x \to b^-} f(x) = f(b)$
Properties of Continuous Functions
If $f$ and $g$ are continuous at $x = a$, then:
| Operation | Continuity |
|---|---|
| $f(x) + g(x)$ | Continuous at $a$ |
| $f(x) - g(x)$ | Continuous at $a$ |
| $f(x) \cdot g(x)$ | Continuous at $a$ |
| $\frac{f(x)}{g(x)}$ | Continuous at $a$ (if $g(a) \neq 0$) |
| $c \cdot f(x)$ | Continuous at $a$ |
| $f(g(x))$ | Continuous at $a$ (if $g$ continuous at $a$) |
Always Continuous Functions
These functions are continuous everywhere in their domain:
- Polynomial functions: $x^n$, $ax^2 + bx + c$
- Exponential: $e^x$, $a^x$
- Logarithmic: $\ln x$ (for $x > 0$)
- Trigonometric: $\sin x$, $\cos x$, $\tan x$ (except where undefined)
- Absolute value: $|x|$
Important Theorems
Intermediate Value Theorem (IVT)
Statement: If $f$ is continuous on $[a, b]$ and $k$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c \in (a, b)$ such that:
$$\boxed{f(c) = k}$$In simple terms: A continuous function takes on every value between $f(a)$ and $f(b)$.
Application: Finding roots of equations!
To show $f(x) = 0$ has a solution in $(a, b)$:
- Check $f$ is continuous on $[a, b]$
- Show $f(a) \cdot f(b) < 0$ (opposite signs)
- By IVT, there exists $c$ where $f(c) = 0$!
JEE loves this pattern!
Example:
Show that $x^3 - 2x - 5 = 0$ has a root in $(2, 3)$.
Let $f(x) = x^3 - 2x - 5$.
- $f(2) = 8 - 4 - 5 = -1 < 0$
- $f(3) = 27 - 6 - 5 = 16 > 0$
Since $f$ is continuous (polynomial) and $f(2) < 0 < f(3)$, by IVT there exists $c \in (2, 3)$ where $f(c) = 0$.
Mean Value Theorem (Preview)
If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then:
$$\boxed{f'(c) = \frac{f(b) - f(a)}{b - a} \text{ for some } c \in (a, b)}$$There exists a point where the instantaneous rate of change equals the average rate of change.
(We’ll cover this in depth in Applications of Derivatives)
Checking Continuity: Step-by-Step
For a Point $x = a$
Step 1: Check if $f(a)$ is defined
Step 2: Calculate LHL: $\lim_{x \to a^-} f(x)$
Step 3: Calculate RHL: $\lim_{x \to a^+} f(x)$
Step 4: If LHL = RHL, that’s the limit. Check if it equals $f(a)$.
Step 5: If all match → Continuous. Otherwise → Discontinuous (identify type).
For Piecewise Functions
At boundary points, use the above method. Inside each piece, check if the formula is continuous.
Practice Problems
Level 1: Foundation (NCERT)
Question: Is $f(x) = x^2 + 3x - 1$ continuous at $x = 2$?
Solution: Polynomial functions are continuous everywhere.
Alternatively:
- $f(2) = 4 + 6 - 1 = 9$ ✓
- $\lim_{x \to 2} f(x) = 9$ ✓
- Limit equals value ✓
Continuous at $x = 2$.
Question: Check continuity of $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.
Solution:
- $f(2)$ is undefined (division by zero) ✗
Discontinuous at $x = 2$ (removable — limit exists and equals 4).
Level 2: JEE Main
Question:
$$f(x) = \begin{cases} x^2 - 4 & \text{if } x \neq 2 \\ k & \text{if } x = 2 \end{cases}$$Find $k$ for continuity at $x = 2$.
Solution: For continuity: $\lim_{x \to 2} f(x) = f(2)$
$$\lim_{x \to 2} (x^2 - 4) = 4 - 4 = 0$$So $k = 0$ for continuity.
Question:
$$f(x) = \begin{cases} \frac{\sin 3x}{x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases}$$Find $k$ for continuity at $x = 0$.
Solution:
$$\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot 3 = 1 \cdot 3 = 3$$For continuity, $k = 3$.
Question: Discuss continuity of $f(x) = |x - 1|$ at $x = 1$.
Solution:
$$f(x) = \begin{cases} -(x - 1) = 1 - x & \text{if } x < 1 \\ x - 1 & \text{if } x \geq 1 \end{cases}$$At $x = 1$:
- $f(1) = 0$ ✓
- LHL: $\lim_{x \to 1^-} (1 - x) = 0$ ✓
- RHL: $\lim_{x \to 1^+} (x - 1) = 0$ ✓
- LHL = RHL = $f(1) = 0$ ✓
Continuous at $x = 1$.
Level 3: JEE Advanced
Question: Show that $\cos x = x$ has at least one solution in $(0, \pi/2)$.
Solution: Let $f(x) = \cos x - x$.
Check continuity: Both $\cos x$ and $x$ are continuous, so $f(x)$ is continuous on $[0, \pi/2]$.
Evaluate endpoints:
- $f(0) = \cos 0 - 0 = 1 > 0$
- $f(\pi/2) = \cos(\pi/2) - \pi/2 = 0 - \pi/2 < 0$
Since $f(0) > 0 > f(\pi/2)$, by IVT there exists $c \in (0, \pi/2)$ where $f(c) = 0$.
Therefore, $\cos c = c$ has a solution.
Question:
$$f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2} & \text{if } x < 0 \\ a & \text{if } x = 0 \\ \frac{\sqrt{1 + x} - 1}{x} & \text{if } x > 0 \end{cases}$$Find $a$ for continuity at $x = 0$.
Solution: LHL:
$$\lim_{x \to 0^-} \frac{1 - \cos 2x}{x^2} = \lim_{x \to 0} \frac{1 - \cos 2x}{(2x)^2} \cdot 4 = \frac{1}{2} \cdot 4 = 2$$RHL:
$$\lim_{x \to 0^+} \frac{\sqrt{1 + x} - 1}{x}$$Rationalize:
$$= \lim_{x \to 0} \frac{(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)} = \frac{1}{2}$$LHL = 2, RHL = $\frac{1}{2}$
LHL ≠ RHL → No value of $a$ makes $f$ continuous at $x = 0$!
Question: If $f(x) = \frac{|x - 3|}{x - 3}$, discuss continuity at $x = 3$.
Solution:
$$f(x) = \begin{cases} \frac{-(x - 3)}{x - 3} = -1 & \text{if } x < 3 \\ \text{undefined} & \text{if } x = 3 \\ \frac{x - 3}{x - 3} = 1 & \text{if } x > 3 \end{cases}$$- $f(3)$ is undefined ✗
- LHL = -1, RHL = 1
- LHL ≠ RHL ✗
Discontinuous at $x = 3$ (jump discontinuity, cannot be removed).
Common Mistakes to Avoid
Wrong: If $\lim_{x \to a} f(x)$ exists, then $f$ is continuous at $a$. ✗
Correct: You also need $f(a)$ to exist AND equal the limit!
Example: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$. Limit exists (= 2), but $f(1)$ is undefined → Discontinuous.
Always verify DEL:
- Defined
- Limit Exists
- Limit equals value
Missing even one means discontinuity!
For piecewise functions, the value at the boundary point is determined by the formula that includes that point (usually with $\leq$ or $\geq$).
Don’t confuse which piece defines $f(a)$!
Quick Revision Box
| Function Type | Continuous? | Notes |
|---|---|---|
| Polynomials | Everywhere | No exceptions |
| $\sin x$, $\cos x$ | Everywhere | — |
| $\tan x$, $\sec x$ | Except at $\frac{\pi}{2} + n\pi$ | Vertical asymptotes |
| $e^x$, $a^x$ | Everywhere | — |
| $\ln x$ | For $x > 0$ | Undefined for $x \leq 0$ |
| $\frac{1}{x}$ | Except $x = 0$ | Infinite discontinuity |
| $ | x | $ |
| $\lfloor x \rfloor$ | Except integers | Jump discontinuities |
JEE Strategy Tips
Weightage: Continuity appears in 1-2 questions in JEE Main, 2-3 in JEE Advanced (often combined with differentiability).
Time-Saver: For polynomials, exponentials, and trig functions in their domain, just say “continuous everywhere” — saves 1 minute!
IVT is a JEE favorite for proving roots exist. Template: Check continuity + Check sign change + Apply IVT.
Piecewise functions: 80% of JEE continuity problems! Master the LHL/RHL/f(a) drill.
Common Trap: JEE will give you a piecewise function and ask for the value of a parameter. Always set LHL = RHL = f(a).
Teacher’s Summary
Continuity = No breaks in the graph. You can trace it without lifting your pen.
The three conditions (DEL) are non-negotiable: Defined, Exists, Limit equals value.
Types of discontinuity: Removable (fixable), Jump (LHL ≠ RHL), Infinite (vertical asymptote), Oscillatory (no limit).
Piecewise functions are 80% of JEE problems — drill the LHL/RHL check until it’s automatic.
IVT is your root-finding superpower — continuous function + sign change = guaranteed root.
“Continuity is the bridge between algebra and calculus!”
Related Topics
Within Limits, Continuity & Differentiability
- Introduction to Limits — Foundation for continuity
- Standard Limits — Used in continuity checks
- Differentiability — Differentiability implies continuity
Mathematical Foundations
- Functions Basics — Domain, range, and function behavior
- Piecewise Functions — Commonly tested with continuity
Applications
- Intermediate Value Theorem — Root finding
- Applications of Derivatives — Mean Value Theorem
- Integral Calculus — Continuity required for definite integrals
Physics Connections
- Kinematics — Position functions must be continuous
- Wave Motion — Wave functions are continuous
- Electric Potential — Continuous except at charges