Curriculum

Mathematics Grade XI

Grade 11 Mathematics

0/14

Text lesson

Chapter 12: Limits and Derivatives

Class 11 • Mathematics • Chapter 12

Limits and Derivatives

The first taste of calculus: getting close to a value, and measuring how fast things change.

CBSE / NCERT Aligned

School Revise

Chapter Roadmap

Idea of a Limit • Algebra of Limits • Limits of Polynomials • Standard Limits • Derivative From First Principles • Rules of Differentiation • Key Results • Applications

Why Limits and Derivatives Matter

Calculus is the mathematics of change, and this chapter is your first step into it. A limit asks what value a function heads towards as the input gets closer and closer to some number, even if it never quite arrives. A derivative then uses limits to measure the exact rate at which a quantity is changing, such as the speed of a car at one instant or the slope of a curve at one point.

These two ideas power almost all of modern science and engineering. Speed, acceleration, growth rates, and the slope of a graph are all derivatives. In this chapter you will build an intuitive feel for limits, learn the rules for working them out, meet two standard limits that appear everywhere, and then define the derivative from first principles before using quick rules to differentiate common functions.

Key idea

A limit is the value a function approaches as x approaches a chosen number. The derivative f′(x) is a special limit that gives the slope of the curve, and equals limₕ→₀ [f(x + h) − f(x)] / h.

Key Terms You Must Know

Term

Meaning

Example

Limit

The value a function f(x) approaches as x approaches a number a.

limₓ→₂ (x + 3) = 5

Algebra of limits

Limits add, subtract, multiply and divide like ordinary numbers.

lim(f + g) = lim f + lim g

Indeterminate 0/0

A form that needs more work, such as factoring, before the limit is found.

(x² − 9)/(x − 3) at x = 3

Standard limit

A known limit used as a building block.

limₓ→₀ (sin x)/x = 1

Derivative

The instantaneous rate of change, written f′(x) or dy/dx.

slope of the curve at a point

First principles

Finding a derivative straight from the limit definition.

f′(x) = limₕ→₀ [f(x+h) − f(x)]/h

Differentiation

The process of finding a derivative.

d/dx (x²) = 2x

Core Concepts, Step by Step

1. The Intuitive Idea of a Limit

A limit describes where a function is heading. We write limₓ→ₐ f(x) = L to mean that as x gets closer and closer to a, the value f(x) gets closer and closer to L. The point a itself need not be in the picture; what matters is the value the function is approaching from both sides. The table below shows the values of f(x) = x² as x closes in on 2 from below and from above; the outputs close in on 4.

Values of f(x) = x² as x approaches 2

f(x) = x²

1.9

3.61

1.99

3.9601

1.999

3.996001

2.001

4.004001

2.01

4.0401

2.1

4.41

2. The Algebra of Limits

Limits behave just like ordinary arithmetic, which makes them easy to handle. If lim f(x) and lim g(x) both exist, then the limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and the limit of a quotient is the quotient of the limits (provided the bottom limit is not 0). A constant multiple comes straight out: lim [k f(x)] = k lim f(x). These rules let us break a complicated limit into simple pieces.

3. Limits of Polynomials and Rational Functions

For a polynomial, the limit as x approaches a is found by simply substituting x = a, because polynomials are smooth and unbroken. For a rational function (one polynomial divided by another), substitute first; if the bottom is not 0 you are done. If substitution gives the indeterminate form 0/0, it means a common factor is hiding, so factorise the top and bottom, cancel the common factor, and then substitute. This single technique solves most rational limits in the chapter.

4. Two Standard Limits

Two limits appear so often that they are worth knowing by heart. The first is the trigonometric limit limₓ→₀ (sin x)/x = 1, where x is measured in radians. The second is the algebraic limit limₓ→ₐ (xⁿ − aⁿ)/(x − a) = n aⁿ₋¹. Many harder limits are solved by rearranging them until one of these standard forms appears, then replacing it with its known value.

5. The Derivative From First Principles

The derivative of a function f at a point measures how fast f is changing there, and equals the slope of the curve at that point. By definition, f′(x) = limₕ→₀ [f(x + h) − f(x)] / h. The fraction is the slope of the line joining two nearby points on the curve; as h shrinks to 0, that line becomes the tangent, and its slope is the derivative. Working a derivative out straight from this definition is called using first principles.

A curve: as x → a, f(x) → L, and the tangent at P has slope f ′(a)

Curve approaching a limit with a tangent line at P

6. Rules of Differentiation

Once a few derivatives are known, rules let us differentiate quickly without first principles every time. The power rule gives d/dx (xⁿ) = n xⁿ₋¹, the derivative of a constant is 0, and derivatives add term by term. For products, (uv)′ = u′v + uv′, and for quotients, (u/v)′ = (u′v − uv′)/v². The standard trigonometric derivatives are d/dx (sin x) = cos x and d/dx (cos x) = −sin x. The table below lists the ones used most often.

Standard derivatives to know

Function f(x)

Derivative f′(x)

a constant c

0

xⁿ

n xⁿ₋¹

sin x

cos x

cos x

−sin x

k f(x)

k f′(x)

u + v

u′ + v′

Key Results & Proofs

Three results show where the differentiation rules come from. Each is built straight from the definition of a limit or a derivative.

Result 1: The Power Rule

Statement. For a positive whole number n, the derivative of xⁿ is n xⁿ₋¹.

Proof

Use the derivative as a limit and the standard limit for (xⁿ − aⁿ)/(x − a).

f′(a)

=

limₓ→ₐ [f(x) − f(a)] / (x − a)

derivative as a limit (point form)

f′(a)

=

limₓ→ₐ (xⁿ − aⁿ)/(x − a)

put f(x) = xⁿ

f′(a)

=

n aⁿ₋¹

apply the standard limit (xⁿ−aⁿ)/(x−a) = n aⁿ₋¹

This holds at every point a, so f′(x) = n xⁿ₋¹. – rename a as x

So d/dx (x₅) = 5x⁴, and d/dx (x) = 1, since the power 1 gives 1·x⁰ = 1.

Result 2: Constant and Sum Rules

Statement. The derivative of a constant is 0, and the derivative of a sum is the sum of the derivatives.

Proof

Apply the first-principles definition to a constant and to a sum.

For a constant f(x) = c, the difference f(x+h) − f(x) = c − c = 0. – a constant never changes

f′(x)

=

limₕ→₀ 0/h = 0

so the derivative is 0

For a sum, [f(x+h) + g(x+h)] − [f(x) + g(x)] splits into two fractions. – group f-terms and g-terms

(f + g)′

=

f′(x) + g′(x)

the limit of a sum is the sum of limits

With the power rule, this lets us differentiate any polynomial term by term.

Result 3: Derivative of sin x

Statement. The derivative of sin x is cos x.

Proof

Expand sin(x + h), then use the two standard limits.

f′(x)

=

limₕ→₀ [sin(x + h) − sin x] / h

first principles

Expand sin(x + h) = sin x cos h + cos x sin h. – compound angle formula

f′(x)

=

limₕ→₀ [sin x (cos h − 1) + cos x sin h] / h

group the terms

As h → 0, (sin h)/h → 1 and (cos h − 1)/h → 0. – standard limits

f′(x)

=

cos x

only the cos x term survives

A similar argument gives d/dx (cos x) = −sin x.

Worked Examples

Example 1

Question: Evaluate limₓ→₂ (x² + 3).

▶ Show full working

A polynomial is smooth, so just substitute x = 2.

limₓ→₂ (x² + 3)

=

2² + 3

substitute x = 2

=

4 + 3 = 7

simplify

Answer: The limit is 7.

Example 2

Question: Evaluate limₓ→₃ (x² − 9)/(x − 3).

▶ Show full working

Substituting gives 0/0, so factorise and cancel first.

At x = 3 the form is 0/0, so a factor is hidden. – check the form

(x² − 9)/(x − 3)

=

(x − 3)(x + 3)/(x − 3)

factorise the top

=

x + 3

cancel (x − 3)

limₓ→₃

=

3 + 3 = 6

now substitute x = 3

Answer: The limit is 6.

Example 3

Question: Evaluate limₓ→₀ (sin x)/x.

▶ Show full working

This is one of the two standard limits.

limₓ→₀ (sin x)/x

=

1

standard limit (x in radians)

Answer: The limit is 1.

Example 4

Question: Evaluate limₓ→₀ (sin 3x)/x.

▶ Show full working

Make the inside of sin match the denominator, then use the standard limit.

(sin 3x)/x

=

3 · (sin 3x)/(3x)

multiply and divide by 3

As x → 0, (sin 3x)/(3x) → 1. – standard limit

limₓ→₀

=

3 × 1 = 3

multiply

Answer: The limit is 3.

Example 5

Question: Evaluate limₓ→₂ (x⁵ − 32)/(x − 2).

▶ Show full working

This matches the standard limit (xⁿ − aⁿ)/(x − a) with n = 5, a = 2.

Note 32 = 2⁵, so a = 2 and n = 5. – spot the standard form

limₓ→₂

=

n aⁿ₋¹ = 5 × 2⁴

apply the standard limit

=

5 × 16 = 80

evaluate

Answer: The limit is 80.

Example 6

Question: Find the derivative of f(x) = x² from first principles.

▶ Show full working

Use f′(x) = limₕ→₀ [f(x + h) − f(x)] / h.

f′(x)

=

limₕ→₀ [(x + h)² − x²] / h

first principles

=

limₕ→₀ [x² + 2xh + h² − x²] / h

expand the square

=

limₕ→₀ [2xh + h²] / h

simplify the top

=

limₕ→₀ (2x + h)

cancel h

=

2x

let h → 0

Answer: f′(x) = 2x.

Example 7

Question: Differentiate y = x⁷.

▶ Show full working

Use the power rule d/dx (xⁿ) = n xⁿ₋¹.

dy/dx

=

7 x⁷₋¹

power rule with n = 7

dy/dx

=

7x⁶

simplify

Answer: dy/dx = 7x⁶.

Example 8

Question: Differentiate y = 3x² + 5x − 2.

▶ Show full working

Differentiate term by term using the power and constant rules.

dy/dx

=

3(2x) + 5(1) − 0

differentiate each term

dy/dx

=

6x + 5

simplify

Answer: dy/dx = 6x + 5.

Example 9

Question: Differentiate y = sin x + cos x.

▶ Show full working

Use the standard trigonometric derivatives.

dy/dx

=

cos x + (−sin x)

derivative of sin is cos, of cos is −sin

dy/dx

=

cos x − sin x

tidy up

Answer: dy/dx = cos x − sin x.

Example 10

Question: Differentiate y = x² sin x.

▶ Show full working

This is a product, so use (uv)′ = u′v + uv′ with u = x² and v = sin x.

u′

=

2x

derivative of x²

v′

=

cos x

derivative of sin x

dy/dx

=

(2x)(sin x) + (x²)(cos x)

apply the product rule

dy/dx

=

2x sin x + x² cos x

tidy up

Answer: dy/dx = 2x sin x + x² cos x.

Example 11

Question: Differentiate y = (2x + 1)/(x − 1).

▶ Show full working

This is a quotient, so use (u/v)′ = (u′v − uv′)/v².

u′

=

2

derivative of 2x + 1

v′

=

1

derivative of x − 1

dy/dx

=

[2(x − 1) − (2x + 1)(1)] / (x − 1)²

apply the quotient rule

dy/dx

=

[2x − 2 − 2x − 1] / (x − 1)²

expand the top

dy/dx

=

−3 / (x − 1)²

simplify

Answer: dy/dx = −3/(x − 1)².

Example 12

Question: Find the derivative of f(x) = 1/x from first principles.

▶ Show full working

Use the definition and combine the fractions over a common denominator.

f′(x)

=

limₕ→₀ [1/(x + h) − 1/x] / h

first principles

=

limₕ→₀ [x − (x + h)] / [h · x(x + h)]

common denominator on top

=

limₕ→₀ (−h) / [h · x(x + h)]

simplify the top

=

limₕ→₀ (−1)/[x(x + h)]

cancel h

=

−1/x²

let h → 0

Answer: f′(x) = −1/x².

Where You Meet This in Real Life

Speed and motion

Speed is the derivative of distance with respect to time, and acceleration is the derivative of speed, so calculus describes all motion.

Rates of change

How fast a population grows, a tank fills, or a temperature falls are all derivatives of one quantity with respect to another.

Slopes and design

The derivative gives the slope of a curve at a point, used to design smooth roads, ramps and curved surfaces.

Economics

Marginal cost and marginal revenue, which guide business decisions, are derivatives of cost and revenue.

Physics and engineering

From electric currents to the cooling of an engine, limits and derivatives are the language used to model change precisely.

Practice Sets A to D

Practice Set A – Basics

A1. Evaluate limₓ→₁ (3x + 2).

▶ Reveal full working

Substitute x = 1.

limₓ→₁ (3x + 2)

=

3(1) + 2 = 5

substitute

Answer: 5.

A2. Evaluate limₓ→₀ (x² + 5).

▶ Reveal full working

Substitute x = 0.

limₓ→₀ (x² + 5)

=

0 + 5 = 5

substitute

Answer: 5.

A3. Differentiate y = x⁴.

▶ Reveal full working

Use the power rule.

dy/dx

=

4x³

power rule with n = 4

Answer: 4x³.

A4. Differentiate the constant function y = 7.

▶ Reveal full working

The derivative of a constant is 0.

dy/dx

=

0

constant rule

Answer: 0.

Practice Set B – Conceptual

B1. What does limₓ→ₐ f(x) = L mean in plain words?

▶ Reveal full working

Describe how the output behaves.

As x gets closer and closer to a, f(x) gets closer and closer to L.

The value at a itself need not matter.

Answer: As x approaches a, f(x) approaches L.

B2. State the value of limₓ→₀ (sin x)/x.

▶ Reveal full working

Recall the standard limit (x in radians).

This is a standard limit equal to 1.

Answer: 1.

B3. Write the first-principles definition of the derivative.

▶ Reveal full working

It is a limit of a difference quotient.

f′(x)

=

limₕ→₀ [f(x + h) − f(x)]/h

the definition

Answer: f′(x) = limₕ→₀ [f(x + h) − f(x)]/h.

B4. Is the derivative of a sum equal to the sum of the derivatives?

▶ Reveal full working

Recall the sum rule.

Yes, differentiation can be done term by term.

(f + g)′ = f′ + g′.

Answer: Yes, by the sum rule.

Practice Set C – Application / Numerical

C1. Evaluate limₓ→₂ (x² − 4)/(x − 2).

▶ Reveal full working

It is 0/0, so factorise and cancel.

(x² − 4)/(x − 2)

=

(x − 2)(x + 2)/(x − 2)

factorise

=

x + 2

cancel

limₓ→₂

=

2 + 2 = 4

substitute

Answer: 4.

C2. Evaluate limₓ→₀ (sin 5x)/x.

▶ Reveal full working

Match the inside of sin to the denominator.

(sin 5x)/x

=

5 · (sin 5x)/(5x)

multiply and divide by 5

limₓ→₀

=

5 × 1 = 5

standard limit

Answer: 5.

C3. Differentiate y = 4x³ − 2x + 7.

▶ Reveal full working

Differentiate term by term.

dy/dx

=

4(3x²) − 2(1) + 0

power and constant rules

dy/dx

=

12x² − 2

simplify

Answer: 12x² − 2.

C4. Differentiate y = x³ cos x.

▶ Reveal full working

Use the product rule with u = x³, v = cos x.

u′

=

3x²

derivative of x³

v′

=

−sin x

derivative of cos x

dy/dx

=

3x² cos x + x³(−sin x)

product rule

dy/dx

=

3x² cos x − x³ sin x

tidy up

Answer: 3x² cos x − x³ sin x.

Practice Set D – HOTS / Word Problems

D1. Find the derivative of f(x) = x³ from first principles.

▶ Reveal full working

Expand (x + h)³ and simplify before letting h → 0.

f′(x)

=

limₕ→₀ [(x+h)³ − x³]/h

first principles

=

limₕ→₀ [3x²h + 3xh² + h³]/h

expand and subtract x³

=

limₕ→₀ (3x² + 3xh + h²)

cancel h

=

3x²

let h → 0

Answer: f′(x) = 3x².

D2. Evaluate limₓ→₃ (x⁴ − 81)/(x − 3).

▶ Reveal full working

Use the standard limit with n = 4, a = 3 (since 81 = 3⁴).

limₓ→₃

=

n aⁿ₋¹ = 4 × 3³

apply the standard limit

=

4 × 27 = 108

evaluate

Answer: 108.

D3. Differentiate y = (x² + 1)/x.

▶ Reveal full working

Split the fraction first, then differentiate.

y

=

(x² + 1)/x = x + 1/x

split the fraction

dy/dx

=

1 − 1/x²

differentiate x and 1/x

Answer: 1 − 1/x².

D4. Evaluate limₓ→₀ (sin 2x)/(sin 3x).

▶ Reveal full working

Turn each part into a standard limit form.

(sin 2x)/(sin 3x)

=

[(sin 2x)/(2x) · 2x] / [(sin 3x)/(3x) · 3x]

insert matching factors

Each (sin kx)/(kx) → 1 as x → 0. – standard limits

limₓ→₀

=

2x/3x = 2/3

what remains

Answer: 2/3.

Chapter Summary

Everything in One Glance

Limit

limₓ→ₐ f(x) = L means f(x) approaches L as x approaches a.

Algebra of Limits

Limits of sums, products and quotients split into limits of the parts.

Rational Limits

Substitute; if you get 0/0, factorise, cancel, then substitute again.

Standard Limits

limₓ→₀ (sin x)/x = 1 and limₓ→ₐ (xⁿ − aⁿ)/(x − a) = n aⁿ₋¹.

Derivative

f′(x) = limₕ→₀ [f(x + h) − f(x)]/h, the slope of the curve.

Rules

Power rule n xⁿ₋¹; constant gives 0; sum, product (u′v + uv′) and quotient rules; sin→cos, cos→−sin.

Are You Exam-Ready?

8-Point Exam Quick-Check

Evaluate limₓ→₂ (4x − 1).

Evaluate limₓ→₃ (x² − 9)/(x − 3).

State the value of limₓ→₀ (sin x)/x.

Differentiate y = x⁶.

Differentiate y = 5x² − 3x + 4.

Find the derivative of sin x.

Differentiate y = x sin x using the product rule.

Find the derivative of x² from first principles.

School Revise Virtual Lab

Practice the concepts in this chapter with interactive simulations and visual tools.

Open the Virtual Lab →

Class 11 Mathematics Chapter 12: Limits and Derivatives, Complete Notes and Practice

These free Class 11 Maths Chapter 12 Limits and Derivatives notes follow the latest NCERT 2026-27 syllabus and give complete, exam-ready coverage for board exams and competitive foundations. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.