Class 11 Maths Chapter 12: Limits and Derivatives

Limits and Derivatives is Chapter 12 of NCERT Class 11 Maths. It is your first step into calculus: you learn what it means for a function to approach a value, how to evaluate limits of algebraic and trigonometric functions, and how the derivative measures the rate of change of a function. Watch the 9 video lessons below, then use the notes, formulas and solved examples to practise.

What you will learn

The intuitive idea of a derivative as an instantaneous rate of change
Left hand and right hand limits and when a limit exists
The algebra of limits and methods for polynomial and rational functions
Standard limits of algebraic and trigonometric functions
The derivative from first principles and the product and quotient rules

Video Lessons

Limits and Derivatives - Part 1

Limits and Derivatives - Part 2

Limits and Derivatives - Part 3

Limits and Derivatives - Part 4

Limits and Derivatives - Part 5

Limits and Derivatives - Part 6

Limits and Derivatives - Part 7

Limits and Derivatives - Part 8

Limits and Derivatives - Part 9

Key Concepts and Formulas

Intuitive Idea of a Derivative

If a ball's position s(t) is known, its average speed over a small time interval is the change in distance divided by the change in time. As that interval shrinks towards zero, the average speed settles to the speed at that exact instant. This settling value is a limit, and it is the derivative of s(t).

Average rate of change from x to x + h:
(f(x + h) − f(x))/h
Instantaneous rate = its value as h → 0

Secant lines approaching the tangent to y = x² + 3x at x = 1, Class 11 Maths Chapter 12 Limits and Derivatives — As h shrinks, secant slopes approach 5

Geometrically, the derivative at a point is the slope of the tangent to the curve there

Understanding Limits

We write lim x→a f(x) = L when f(x) gets as close as we like to L for x close enough to a (but not equal to a). The function need not even be defined at x = a.

LHL = lim x→a⁻ f(x)  (x less than a)
RHL = lim x→a⁺ f(x)  (x greater than a)
Limit exists ⇔ LHL = RHL

When the limit exists, it equals the common value of LHL and RHL
For piecewise functions, always check both sides at the point where the rule changes

Algebra of Limits

If lim x→a f(x) and lim x→a g(x) both exist, limits behave nicely with the four basic operations.

lim (f ± g) = lim f ± lim g
lim (f × g) = lim f × lim g
lim (f/g) = lim f / lim g,  lim g ≠ 0
lim (k × f) = k × lim f
Polynomial: lim x→a p(x) = p(a)

For a rational function p(x)/q(x), substitute if q(a) ≠ 0
If substitution gives 0/0, factorise and cancel the common factor (x − a) or rationalise

Standard Limits

These results are proved in the chapter and should be memorised. Many questions reduce to one of them after a little rearranging.

lim x→a (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹
lim x→0 (sin x)/x = 1
lim x→0 (1 − cos x)/x = 0
lim x→0 (tan x)/x = 1
lim x→0 (sin ax)/(bx) = a/b

The trigonometric limits hold only when x is measured in radians

Derivative from First Principles

The derivative of f at a point is defined as a limit. Using this definition directly is called finding the derivative from first principles.

f′(x) = lim h→0 (f(x + h) − f(x))/h
f′(a) = lim h→0 (f(a + h) − f(a))/h

The derivative exists only when this limit exists
Other notations: dy/dx, d/dx (f(x)), y′

Rules and Standard Derivatives

Once the definition gives the basic results, the rules let you differentiate combinations quickly.

Sum rule: (u ± v)′ = u′ ± v′
Product rule: (uv)′ = u′v + uv′
Quotient rule: (u/v)′ = (u′v − uv′)/v², v ≠ 0
d/dx (xⁿ) = nxⁿ⁻¹
d/dx (constant) = 0
d/dx (sin x) = cos x
d/dx (cos x) = −sin x
d/dx (tan x) = sec²x

Solved Examples

Example 1: Evaluate lim x→2 (x³ + 3x − 5).

Polynomial, so substitute x = 2
= 2³ + 3(2) − 5
= 8 + 6 − 5
∴ Limit = 9

Example 2: Evaluate lim x→3 (x² − 9)/(x² − 5x + 6).

At x = 3 we get 0/0, so factorise
= (x − 3)(x + 3) / ((x − 3)(x − 2))
= (x + 3)/(x − 2)
= (3 + 3)/(3 − 2) = 6/1
∴ Limit = 6

Graph with a hole at (3, 6) showing left hand and right hand limits for Class 11 Limits and Derivatives — Both sides approach 6, so the limit is 6

Example 3: Evaluate lim x→2 (x⁵ − 32)/(x − 2).

32 = 2⁵, so the limit is (x⁵ − 2⁵)/(x − 2)
Use naⁿ⁻¹ with n = 5, a = 2
= 5 × 2⁴ = 5 × 16
∴ Limit = 80

Example 4: Evaluate lim x→0 (sin 6x)/(tan 2x).

= (sin 6x/6x) × (2x/tan 2x) × (6x/2x)
sin 6x/6x → 1 and 2x/tan 2x → 1
= 1 × 1 × 6/2
∴ Limit = 3

Example 5: Find the derivative of f(x) = x² + 3x from first principles.

f(x + h) = x² + 2xh + h² + 3x + 3h
f(x + h) − f(x) = 2xh + h² + 3h
Divide by h: 2x + h + 3
Let h → 0
∴ f′(x) = 2x + 3

Example 6: Differentiate y = (2x + 1)(x² − 4) using the product rule.

u = 2x + 1, u′ = 2
v = x² − 4, v′ = 2x
y′ = 2(x² − 4) + (2x + 1)(2x)
= 2x² − 8 + 4x² + 2x
∴ y′ = 6x² + 2x − 8

Example 7: Differentiate y = (x² + 1)/(x − 2) using the quotient rule.

u = x² + 1, u′ = 2x
v = x − 2, v′ = 1
y′ = (2x(x − 2) − (x² + 1)(1))/(x − 2)²
= (2x² − 4x − x² − 1)/(x − 2)²
∴ y′ = (x² − 4x − 1)/(x − 2)²

Common Mistakes to Avoid

Cancelling (x − a) before checking that the form is really 0/0. Always try direct substitution first.
Using the trigonometric standard limits with angles in degrees. They hold only when x is in radians.
Getting the order wrong in the quotient rule. It is u′v − uv′ in the numerator, not uv′ − u′v.
Concluding that a limit exists after checking only one side for piecewise functions.

Quick Revision

lim x→a (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹
lim x→0 sin x / x = 1
lim x→0 (1 − cos x)/x = 0
f′(x) = lim h→0 (f(x + h) − f(x))/h
(uv)′ = u′v + uv′
(u/v)′ = (u′v − uv′)/v²
d/dx xⁿ = nxⁿ⁻¹
d/dx sin x = cos x, d/dx cos x = −sin x
d/dx tan x = sec²x