JEE Mains Maths · Limits and Continuity

Standard Limits and Algebraic Forms

Limits of the form 0/0 settled by the standard trigonometric, exponential and logarithmic limits, or by factorising and rationalising.

All Limits and Continuity notes JEE Mains Maths strategy

Why this matters

Twenty-four PYQs, twenty-one of them multiple choice. Thirteen reduce to the standard limits such as sin x / x and (eˣ − 1)/x; eleven cancel a common factor or rationalise a surd first. Two ideas cover the page.

Concept 1 of 2: The standard limits

Most 0/0 limits near 0 are built from a few standard ones. Rewrite the expression so each small quantity appears in its standard shape —

\frac{\sin(3x)}{3x}

\frac{e^{2x}-1}{2x}

— and each shape tends to 1. What is left is a plain ratio of constants.

Definition

$\lim_{x\to0}\frac{\sin x}x=\lim_{x\to0}\frac{\tan x}x=1$ .
$\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$ .
$\lim_{x\to0}\frac{e^x-1}x=1$ , $\lim_{x\to0}\frac{a^x-1}x=\ln a$ , $\lim_{x\to0}\frac{\ln(1+x)}x=1$ .
$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$ .

Standard limits

\lim_{x\to0}\frac{\sin x}x=1,\quad\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12,\quad\lim_{x\to0}\frac{e^x-1}x=1

Worked example

Find

\lim_{x\to0}\frac{\sin3x}{\tan5x}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q64Moderate

Example 1 · Limits and Continuity · Standard Limits and Algebraic Forms

The value of

\underset{x\rightarrow 0}{\lim} \frac{\log_{e}\left( sec(ex) \cdot sec\left( e^{2}x \right)\cdot \ldots \cdot sec\left( e^{10}x \right) \right)}{e^{2}-e^{2\cos x}}

is equal to

Match the argument exactly

\frac{\sin3x}x

tends to 3, not 1: the angle and the denominator must be the same before the standard limit applies. Multiply and divide to make them match.

Drill 12 more on the standard limits

Concept 2 of 2: Factorise or rationalise

A 0/0 form at

x=a

means

x-a

is hiding in both numerator and denominator. Factorise polynomials to cancel it; for a surd, multiply by the conjugate so the difference of squares brings the factor out. After cancelling, substitute.

Definition

Polynomial 0/0 at $a$ : both have the factor $x-a$ ; cancel it.
Surd: multiply by the conjugate, $(\sqrt p-\sqrt q)(\sqrt p+\sqrt q)=p-q$ .
After cancelling, substitute directly.

Conjugate

\sqrt p-\sqrt q=\frac{p-q}{\sqrt p+\sqrt q}

Worked example

Find

\lim_{x\to2}\frac{x^3-8}{x^2-4}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q155Moderate

Example 2 · Limits and Continuity · Standard Limits and Algebraic Forms

The value of

\lim_{x \rightarrow 1} \frac{\left( x^{2}- 1 \right)\sin^{2}(\pi x)}{x^{4}- 2x^{3}+ 2x - 1}

is equal to:

Rationalise the right part

Multiply by the conjugate of the surd that causes the 0, and keep the other factor. Rationalising a denominator that is not zero only adds work.

Drill 10 more on factorise or rationalise

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)

The standard limits
Standard limits
$\lim_{x\to0}\frac{\sin x}x=1,\quad\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12,\quad\lim_{x\to0}\frac{e^x-1}x=1$
Factorise or rationalise
Conjugate
$\sqrt p-\sqrt q=\frac{p-q}{\sqrt p+\sqrt q}$

Watch out for (2)

Match the argument exactly
→ The standard limits
Rationalise the right part
→ Factorise or rationalise

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 24 questions