JEE Mains Maths · Limits and Continuity

Counting Discontinuities and Non-Differentiability

Counting the points where a function breaks or has a corner: jumps of the greatest integer function, compositions, and functions defined as a limit.

All Limits and Continuity notes JEE Mains Maths strategy

Why this matters

Eighteen PYQs, half of them numerical answers, and four from 2026. Nine count where a greatest-integer expression jumps or where a modulus makes a corner; four compose two functions; five define f as a limit in n and ask where the result breaks. Three ideas cover the page.

Concept 1 of 3: Jumps and corners

[g(x)]

jumps wherever

g(x)

crosses an integer, so list those crossings in the interval — unless a factor multiplying it is zero there, which can cancel the jump. A modulus

|g(x)|

has a corner where

g

changes sign with non-zero slope; a function is not differentiable at a jump or a corner.

Definition

$[g(x)]$ : possible breaks where $g(x)$ is an integer.
Check each: a factor that vanishes there can make the product continuous.
$|g(x)|$ : corner at a simple zero of $g$ ; smooth at a double zero.
Not continuous means not differentiable.

Where the greatest integer jumps

[g(x)]\ \text{can break only where}\ g(x)\in\mathbb Z

Worked example

Where is

[x^2]

discontinuous on

(0,2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q75Moderate

Example 1 · Limits and Continuity · Counting Discontinuities and Non-Differentiability

The number of points in the interval [2,4], at which the function

f(x) = \left\lbrack x^{2} - x - \frac{1}{2} \right\rbrack

, where

\lbrack \cdot \rbrack

denotes the greatest integer function, is discontinuous, is

\_\_\_\_

A zero factor can hide a jump

x[x]

at 0 the jump of

[x]

is multiplied by 0, so the product is continuous. Test each candidate point; do not just count the integer crossings.

Drill 8 more on jumps and corners

Concept 2 of 3: Compositions

f(g(x))

can break where

g

breaks, or where

g(x)

lands on a point where

f

breaks. So find the bad points of

f

, solve

g(x)=

each of them, and add the bad points of

g

. Then check each candidate.

Definition

Candidates: breaks of $g$ , and $x$ with $g(x)$ at a break of $f$ .
Continuous inside and outside gives a continuous composition.
Check each candidate by the left and right limits.

Candidate points

\{x:g\ \text{breaks}\}\cup\{x:g(x)\ \text{is a break of}\ f\}

Worked example

Where is

[\sin x]

discontinuous in

(0,2\pi)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q75Moderate

Example 2 · Limits and Continuity · Counting Discontinuities and Non-Differentiability

Let

f(x) = \left\{ \begin{matrix} x^{3} + 8 & x < 0 \\ x^{2} - 4 & x \geq 0 \end{matrix} \right.

and

g(x) = \left\{ \begin{matrix} (x - 8)^{1/3}; & x < 0 \\ (x + 4)^{1/2}; & x \geq 0 \end{matrix} \right.

Then the number of points, where the function gof is discontinuous, is

\_\_\_\_

Also where the inside lands on a bad point

Checking only where

g

breaks misses the points where

g(x)

equals a break of

f

. Solve

g(x)=

each bad point of

f

Drill 3 more on compositions

Concept 3 of 3: Functions defined as a limit

When

f(x)=\lim_{n\to\infty}

of an expression with

x^n

, split by

|x|

: for

|x|<1

x^n\to0

; for

|x|>1

it dominates; at

x=\pm1

compute directly. The result is a piecewise function, and its breaks are usually at

x=\pm1

Definition

$|x|<1$ : $x^{n}\to0$ .
$|x|>1$ : divide by the dominant power.
$x=1$ and $x=-1$ : substitute.
Then check continuity at the boundaries.

Powers in the limit

\lim_{n\to\infty}x^{2n}=\begin{cases}0,&|x|<1\\1,&|x|=1\\\infty,&|x|>1\end{cases}

Worked example

f(x)=\lim_{n\to\infty}\frac{x^{2n}}{1+x^{2n}}

. Where is

f

discontinuous?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q55Moderate

Example 3 · Limits and Continuity · Counting Discontinuities and Non-Differentiability

Let

f(x) =\lim_{\theta\rightarrow 0} \left( \frac{\cos\pi x-x^{\left( \frac{2}{\theta} \right)}\sin(x- 1)}{1 +x^{\left( \frac{2}{\theta} \right)}(x- 1)} \right),x\in R

. Consider the following two statements : (I)

f(x)

is discontinous at

x = 1

. (II)

f(x)

is continous at

x = - 1

. Then,

Compute the boundary separately

x=\pm1

the power

x^{2n}

is exactly 1, giving a third value. Leaving the boundary out misses the break.

Drill 4 more on functions defined as a limit

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Jumps and corners
Where the greatest integer jumps
$[g(x)]\ \text{can break only where}\ g(x)\in\mathbb Z$
Compositions
Candidate points
$\{x:g\ \text{breaks}\}\cup\{x:g(x)\ \text{is a break of}\ f\}$
Functions defined as a limit
Powers in the limit
$\lim_{n\to\infty}x^{2n}=\begin{cases}0,&|x|<1\\1,&|x|=1\\\infty,&|x|>1\end{cases}$

Watch out for (3)

A zero factor can hide a jump
→ Jumps and corners
Also where the inside lands on a bad point
→ Compositions
Compute the boundary separately
→ Functions defined as a limit

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