JEE Mains Maths · Limits and Continuity

Continuity at a Point

A function is continuous at a point when its left limit, right limit and value there all agree; most questions find the constants that make that happen.

All Limits and Continuity notes JEE Mains Maths strategy

Why this matters

Twenty PYQs, sixteen of them multiple choice, and four from 2026. Sixteen give a piecewise function with unknown constants and ask for them from continuity at one point; four use continuity to find a value or to show that an equation has a root. Two ideas cover the page.

Concept 1 of 2: Matching the two sides

At the join of a piecewise rule, compute the limit from each side using that side's formula, and set both equal to the value at the point. Each side often needs a standard limit or a

1^\infty

formula; the equations that result give the constants.

Definition

Continuous at $a$ : $\lim_{x\to a^-}f=\lim_{x\to a^+}f=f(a)$ .
Use each piece's own formula on its own side.
Two constants usually need two equations: left = value and right = value.

Continuity

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

Worked example

f(x)=ax+1

for

x\le1

and

x^2+2

for

x>1

. Find

a

for continuity.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q70Moderate

Example 1 · Limits and Continuity · Continuity at a Point

Let

\lbrack t\rbrack

denote the greatest integer less than or equal to

t

. If the function

f\left( x \right)=\left\{ \begin{matrix} b^{2}\sin\left( \frac{\pi}{2}\left\lbrack \frac{\pi}{2}\left( \cos x + \sin x \right)\cos x \right\rbrack \right) & \ ,x < 0 \\ \frac{\sin x -\frac{1}{2}\sin2x}{x^{3}} & \ ,x > 0 \\ a & \ ,x = 0 \end{matrix} \right.

is continuous at

x = 0

, then

a^{2}+b^{2}

is equal to

Use the right formula on each side

The left limit uses the piece defined for

x<a

, and the value

f(a)

uses whichever piece includes

a

. Mixing them gives a wrong equation.

Drill 15 more on matching the two sides

Concept 2 of 2: Using continuity

Continuity fills in a value a formula leaves undefined: the value must be the limit. It also guarantees roots: a continuous function that changes sign on an interval is zero somewhere inside (the intermediate value theorem).

Definition

Removable gap: define $f(a)=\lim_{x\to a}f(x)$ .
Intermediate value theorem: $f$ continuous on $[a,b]$ , $f(a)f(b)<0$ gives a root in $(a,b)$ .
Sums, products and compositions of continuous functions are continuous.

Intermediate value theorem

f(a)\,f(b)<0\ \Rightarrow\ f(c)=0\ \text{for some}\ c\in(a,b)

Worked example

Show that

x^3+x-1=0

has a root in

(0,1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q59Moderate

Example 2 · Limits and Continuity · Continuity at a Point

Let

f(x) =\left\{ \begin{matrix} \frac{ax^{2}+ 2ax + 3}{4x^{2}+ 4x - 3}, & x \neq -\frac{3}{2},\frac{1}{2} \\ b & x = -\frac{3}{2},\frac{1}{2} \end{matrix} \right.

be continuous at

x = -\frac{3}{2}

. If

fof(x) =\frac{7}{5}

, then

x

is equal to :

No sign change, no guarantee

f(a)

and

f(b)

have the same sign, the theorem says nothing: there may be two roots or none.

Drill 3 more on using continuity

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 (2)

Matching the two sides
Continuity
$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$
Using continuity
Intermediate value theorem
$f(a)\,f(b)<0\ \Rightarrow\ f(c)=0\ \text{for some}\ c\in(a,b)$

Watch out for (2)

Use the right formula on each side
→ Matching the two sides
No sign change, no guarantee
→ Using continuity

Test yourself on Limits and Continuity

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Limits and Continuity chapter test Drill all 20 questions