JEE Mains Maths · Limits and Continuity

Series Expansions and Unknown Constants

Evaluating limits by writing each function as the first few terms of its series, and choosing unknown constants so that a limit exists.

All Limits and Continuity notes JEE Mains Maths strategy

Why this matters

Twenty-two PYQs, seventeen of them multiple choice, and five from 2026. Six evaluate a limit by expanding sin x, cos x, eˣ or ln(1 + x); sixteen ask for constants that make a limit finite, then for the limit itself. Two ideas cover the page.

Concept 1 of 2: Limits by series expansion

Near 0, replace each function by its series up to the power that matters: the lowest power in the denominator. Terms cancel, and the first surviving term decides the limit. This works where standard limits do not, as in

\frac{x-\sin x}{x^3}

Definition

$\sin x=x-\frac{x^3}6+\dots$ , $\cos x=1-\frac{x^2}2+\frac{x^4}{24}-\dots$ .
$e^x=1+x+\frac{x^2}2+\frac{x^3}6+\dots$ , $\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\dots$ .
$\tan x=x+\frac{x^3}3+\dots$ , $(1+x)^n=1+nx+\frac{n(n-1)}2x^2+\dots$ .

Key expansions

\sin x=x-\tfrac{x^3}{6}+\cdots,\quad\cos x=1-\tfrac{x^2}{2}+\cdots,\quad e^x=1+x+\tfrac{x^2}{2}+\cdots

Worked example

Find

\lim_{x\to0}\frac{x-\sin x}{x^3}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q67Moderate

Example 1 · Limits and Continuity · Series Expansions and Unknown Constants

The value of

\lim_{x\rightarrow 0} \left( \frac{x^{2}\sin^{2}x}{x^{2}-\sin^{2}x} \right)

is:

Expand far enough

Stop too early and everything cancels, leaving

\frac00

again. Expand each function to the power of

x

in the denominator.

Drill 5 more on limits by series expansion

Concept 2 of 2: Constants that make a limit finite

If a limit like

\frac{f(x)}{x^n}

is finite, every term of

f

below

x^n

must vanish. Expand

f

, set the coefficients of

1, x, \dots, x^{n-1}

to zero to find the constants, and the coefficient of

x^n

is the limit.

Definition

$\lim\frac{f(x)}{x^n}$ finite: the coefficients of $x^0,\dots,x^{n-1}$ in $f$ are 0.
The limit is then the coefficient of $x^n$ .
One equation per vanishing coefficient: count the unknowns.

Matching coefficients

f(x)=c_0+c_1x+\dots:\ \lim_{x\to0}\frac{f(x)}{x^n}\ \text{finite}\Rightarrow c_0=\dots=c_{n-1}=0

Worked example

Find

a

so that

\lim_{x\to0}\frac{e^{ax}-1-2x}{x^2}

is finite, and the limit.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q66Moderate

Example 2 · Limits and Continuity · Series Expansions and Unknown Constants

\underset{x\rightarrow 2}{\lim} \frac{\sin\left( x^{3}- 5x^{2}+ax+b \right)}{(\sqrt{x- 1}- 1)\log_{e}(x- 1)}=m

, then

a+b+m

is equal to:

Finite is not the same as zero

The lower coefficients must vanish; the

x^n

coefficient need not. Setting it to zero as well answers a different question.

Drill 15 more on constants that make a limit finite

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)

Limits by series expansion
Key expansions
$\sin x=x-\tfrac{x^3}{6}+\cdots,\quad\cos x=1-\tfrac{x^2}{2}+\cdots,\quad e^x=1+x+\tfrac{x^2}{2}+\cdots$
Constants that make a limit finite
Matching coefficients
$f(x)=c_0+c_1x+\dots:\ \lim_{x\to0}\frac{f(x)}{x^n}\ \text{finite}\Rightarrow c_0=\dots=c_{n-1}=0$

Watch out for (2)

Expand far enough
→ Limits by series expansion
Finite is not the same as zero
→ Constants that make a limit finite

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 22 questions