Important Questions Class 11 Maths Chapter 12 Limits and Derivatives 2026–27

Limits and Derivatives introduces students to approaching values, instantaneous rate of change and the derivative of a function.
For CBSE Class 11 Maths, Chapter 12 builds the base for calculus through limits, algebra of limits, standard limits and derivatives from first principle.

Limits and Derivatives is the first calculus chapter in Class 11 Maths. It explains how a function behaves near a point and how the derivative measures change. Students learn left hand limit, right hand limit, standard limits, algebra of limits and basic derivatives.

Use these Important Questions Class 11 Maths Chapter 12 to practise NCERT-based questions for the 2026–27 exams. Start with direct limits, then solve trigonometric limits, derivative of a function, derivatives from first principle and rate of change questions.

Key Takeaways

  • Chapter Name: Chapter 12 in the 2026–27 NCERT Class 11 Maths book is Limits and Derivatives.
  • Core Concepts: Limits, left hand limit, right hand limit and derivative of a function are the main ideas.
  • Formula Practice: Standard limits and algebra of limits are needed for most questions.
  • Exam Focus: Students should practise 1-mark, short answer and long answer questions with steps.

Struggling with limits, derivatives and first principle questions?
Access interactive practice, live doubt-solving sessions, and complete NCERT solutions on the Extramarks Learning App. Sign Up Free

Very Short Answer Questions Class 11 Maths Chapter 12 Limits and Derivatives

These questions test basic concepts from CBSE Class 11 Maths Chapter 12 Limits and Derivatives. Most answers need one formula, one substitution step or one direct definition.

Q1. Evaluate lim x→3 [(x² − 9) / (x − 3)].

Factorise the numerator.

x² − 9 = (x − 3)(x + 3)

So,

lim x→3 [(x² − 9) / (x − 3)]
= lim x→3 [(x − 3)(x + 3) / (x − 3)]
= lim x→3 (x + 3)
= 6

Q2. Evaluate lim x→0 (sin 5x / 3x).

Write the expression using the standard limit.

sin 5x / 3x = (5/3) × (sin 5x / 5x)

So,

lim x→0 (sin 5x / 3x)
= 5/3 × 1
= 5/3

Q3. Evaluate lim x→1 (x³ − 1) / (x − 1).

Use x³ − 1 = (x − 1)(x² + x + 1).

lim x→1 (x³ − 1) / (x − 1)
= lim x→1 (x² + x + 1)
= 1 + 1 + 1
= 3

Q4. Evaluate lim x→2 (x + 10).

This is a polynomial expression, so substitute x = 2.

lim x→2 (x + 10) = 2 + 10 = 12

Q5. Evaluate lim x→1 (x² + x).

Substitute x = 1.

lim x→1 (x² + x)
= 1² + 1
= 2

Q6. State the condition for the existence of lim x→a f(x).

The limit exists when the left hand limit and right hand limit are equal.

That is,

lim x→a− f(x) = lim x→a+ f(x)

Q7. What is the left hand limit?

The left hand limit is the value approached by f(x) when x approaches a from values less than a.

It is written as lim x→a− f(x).

Q8. What is the right hand limit?

The right hand limit is the value approached by f(x) when x approaches a from values greater than a.

It is written as lim x→a+ f(x).

Q9. Find the derivative of f(x) = 3x at x = 2.

For f(x) = 3x, the derivative is 3 for every x.

So,

f′(2) = 3

Q10. Write the derivative of sin x.

The derivative of sin x is cos x.

So,

d/dx (sin x) = cos x

Short Answer Questions Class 11 Maths Chapter 12 Limits and Derivatives

These Class 11 Maths Chapter 12 questions need clear working. Use algebra of limits, standard limits and factorisation wherever direct substitution gives an indeterminate form.

Q11. Evaluate lim x→2 [(x² − 4) / (x − 2)].

Factorise the numerator.

x² − 4 = (x − 2)(x + 2)

So,

lim x→2 [(x² − 4) / (x − 2)]
= lim x→2 [(x − 2)(x + 2) / (x − 2)]
= lim x→2 (x + 2)
= 4

Q12. Evaluate lim x→0 [(1 − cos x) / x].

Use the standard behaviour of trigonometric limits near x = 0.

1 − cos x tends to 0 faster than x near 0.

So,

lim x→0 [(1 − cos x) / x] = 0

Q13. Evaluate lim x→0 (tan x / x).

We know that tan x = sin x / cos x.

So,

lim x→0 (tan x / x)
= lim x→0 [sin x / (x cos x)]
= lim x→0 (sin x / x) × lim x→0 (1 / cos x)
= 1 × 1
= 1

Q14. Evaluate lim x→1 [(x² − 1) / (x² − x)].

Factorise both numerator and denominator.

x² − 1 = (x − 1)(x + 1)
x² − x = x(x − 1)

So,

lim x→1 [(x² − 1) / (x² − x)]
= lim x→1 [(x − 1)(x + 1) / x(x − 1)]
= lim x→1 [(x + 1) / x]
= 2

Q15. Evaluate lim x→2 [(x³ − 8) / (x − 2)].

Use a³ − b³ = (a − b)(a² + ab + b²).

x³ − 8 = (x − 2)(x² + 2x + 4)

So,

lim x→2 [(x³ − 8) / (x − 2)]
= lim x→2 (x² + 2x + 4)
= 4 + 4 + 4
= 12

Q16. Find the derivative of f(x) = x² at x = 3 from first principle.

Using first principle,

f′(3) = lim h→0 [f(3 + h) − f(3)] / h

= lim h→0 [(3 + h)² − 9] / h

= lim h→0 [9 + 6h + h² − 9] / h

= lim h→0 [6h + h²] / h

= lim h→0 (6 + h)

= 6

Q17. Find the derivative of f(x) = 2x² + 3x − 5 at x = 0.

Differentiate the function.

f′(x) = 4x + 3

At x = 0,

f′(0) = 4(0) + 3 = 3

Q18. Find the derivative of f(x) = 2x² + 3x − 5 at x = −1.

Differentiate the function.

f′(x) = 4x + 3

At x = −1,

f′(−1) = 4(−1) + 3
= −4 + 3
= −1

Q19. Verify that f′(0) + 3f′(−1) = 0 for f(x) = 2x² + 3x − 5.

From the previous answers,

f′(0) = 3
f′(−1) = −1

So,

f′(0) + 3f′(−1)
= 3 + 3(−1)
= 3 − 3
= 0

Hence verified.

Q20. Find the derivative of f(x) = x³.

Using the standard derivative formula,

d/dx (x³) = 3x²

So,

f′(x) = 3x²

Long Answer Questions Class 11 Maths Chapter 12 Limits and Derivatives

These Class 11 Maths Limits and Derivatives Important Questions need step-wise solutions. Practise derivatives from first principle, derivative of a function and rate of change questions with proper notation.

Q21. Evaluate lim x→1 [(x⁴ − 1) / (x − 1)].

Factorise x⁴ − 1.

x⁴ − 1 = (x − 1)(x³ + x² + x + 1)

So,

lim x→1 [(x⁴ − 1) / (x − 1)]
= lim x→1 [(x − 1)(x³ + x² + x + 1) / (x − 1)]

= lim x→1 (x³ + x² + x + 1)

= 1 + 1 + 1 + 1

= 4

Q22. Evaluate lim x→0 [(sin 3x) / (sin 5x)].

Rewrite the expression using standard limits.

lim x→0 [(sin 3x) / (sin 5x)]

= lim x→0 [(sin 3x / 3x) × (3x / 5x) × (5x / sin 5x)]

= 1 × 3/5 × 1

= 3/5

Q23. Evaluate lim x→0 [(1 − cos 2x) / x²].

Use 1 − cos 2x = 2sin²x.

So,

lim x→0 [(1 − cos 2x) / x²]
= lim x→0 [2sin²x / x²]

= 2 lim x→0 (sin x / x)²

= 2(1)²

= 2

Q24. Find the derivative of f(x) = x² − 2 at x = 10 from first principle.

Using first principle,

f′(10) = lim h→0 [f(10 + h) − f(10)] / h

Now,

f(10 + h) = (10 + h)² − 2
= 100 + 20h + h² − 2
= 98 + 20h + h²

f(10) = 10² − 2 = 98

So,

f′(10) = lim h→0 [(98 + 20h + h² − 98) / h]

= lim h→0 [(20h + h²) / h]

= lim h→0 (20 + h)

= 20

Q25. Find the derivative of f(x) = x from first principle at x = 1.

Using first principle,

f′(1) = lim h→0 [f(1 + h) − f(1)] / h

= lim h→0 [(1 + h) − 1] / h

= lim h→0 h / h

= 1

So,

f′(1) = 1

Q26. Find the derivative of f(x) = x² − 2x + 1.

Differentiate term by term.

d/dx (x²) = 2x
d/dx (−2x) = −2
d/dx (1) = 0

So,

f′(x) = 2x − 2

Q27. Find the derivative of f(x) = sin x cos x.

Use the product rule.

f(x) = sin x cos x

f′(x) = (sin x)' cos x + sin x (cos x)'

= cos x cos x + sin x(−sin x)

= cos²x − sin²x

So,

f′(x) = cos 2x

Q28. Find the derivative of f(x) = sin²x.

Write sin²x as sin x × sin x.

f′(x) = (sin x)' sin x + sin x (sin x)'

= cos x sin x + sin x cos x

= 2sin x cos x

= sin 2x

So,

f′(x) = sin 2x

Q29. Find the derivative of f(x) = (x − a)(x − b), where a and b are constants.

Expand the expression.

f(x) = (x − a)(x − b)

= x² − ax − bx + ab

= x² − (a + b)x + ab

Differentiate term by term.

f′(x) = 2x − (a + b)

Q30. Find the derivative of f(x) = (x + 1) / (x − 1), where x ≠ 1.

Use the quotient rule.

Let u = x + 1 and v = x − 1.

u′ = 1 and v′ = 1

f′(x) = [v u′ − u v′] / v²

= [(x − 1)(1) − (x + 1)(1)] / (x − 1)²

= [x − 1 − x − 1] / (x − 1)²

= −2 / (x − 1)²

So,

f′(x) = −2 / (x − 1)²

Important Formulas for CBSE Class 11 Maths Chapter 12 Limits and Derivatives

Before solving Important Questions for Class 11 Maths Chapter 12, revise the formulas used in limits and derivatives. These formulas help in direct substitution, factorisation-based limits, trigonometric limits and first principle questions.

Concept Formula Key Variables
Limit of a sum lim x→a [f(x) + g(x)] = lim x→a f(x) + lim x→a g(x) f(x), g(x), a
Standard trigonometric limit lim x→0 sin x / x = 1 x in radians
Derivative from first principle f′(a) = lim h→0 [f(a + h) − f(a)] / h a, h, f(x)

Important Questions Class 11 Maths Chapter-Wise

Chapter No. Chapter Name
Chapter 1 Sets
Chapter 2 Relations and Functions
Chapter 3 Trigonometric Functions
Chapter 4 Complex Numbers and Quadratic Equations
Chapter 5 Linear Inequalities
Chapter 6 Permutations and Combinations
Chapter 7 Binomial Theorem
Chapter 8 Sequences and Series
Chapter 9 Straight Lines
Chapter 10 Conic Sections
Chapter 11 Introduction to Three Dimensional Geometry
Chapter 13 Statistics
Chapter 14 Probability

Q.1 If A(2, 2, 3) and B(13,3,13) are two points, the equation of set of points P such that 3PA = 2PB is

Marks:1

Ans

Let the co-ordinates of point P be (x, y, z).

Now, PA = {(x + 2)2+ (y – 2)2 + (z – 3)2}.

PA2 = (x + 2)2 + (y – 2)2 + (z – 3)2

PB = {(x – 13)2+ (y + 3)2 + (z – 13)2}

PB2 = (x – 13)2 + (y + 3)2 + (z – 13)2

Therefore, from the given condition,

3PA = 2PB 9PA2 = 4PB2

9[(x + 2)2 + (y – 2)2 +(z – 3)2] = 4[(x – 13)2 + (y + 3)2 +(z – 13)2]

On solving this equation we get,

5(x2 + y2 + z2) +140x – 60y – 50z – 1235 = 0.

Q.2 By using distance formula, prove that the three points ( – 4,6,10), (2,4,6) and (14, 0, – 2) are collinear.

Marks:4

Ans

Let points A(4,6,10), B(2,4,6) and C(14,0,2) are given. AB=(2+4)2+(46)2+(610)2=36+4+16=56=214 BC=(142)2+(04)2(26)2 =144+16+64=224=414 AC=(14+4)2+(06)2+(210)2=324+36+144=504=614 AB+BC=214+414=614= AC Thus, AtB and C are collinear.

Q.3 Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, -8) is divided by the XZ-plane.

Marks:2

Ans

LetXZ-planedividesthesegmentjoiningA(4,8,10)andB(6,10,8) atP(x,y,z)in the ratiok:1.Then the coordinates of Pare1(4)+k(6)k+1,1(8)+k(10)k+1,1(10)k(8)k+14+6kk+1,8+10kk+1,108kk+1Since,

PliesonXZ-planeSo,itsY-coordinateiszero,i.e.,8+10kk+1=08+10k=010k=8k=810k=45Therefore, XZ-plane divides AB externallyin the ratio4:5.

Q.4 Name the octants in which the following points lie: (7,4,3) and (5,-3,2).

Points (7,4,3) and (5,-3,2) lie in V and VII Octants respectively.

Marks:1

Ans

Points (7,4,3) and (5,-3,2) lie in V and VII Octants respectively.

Q.5 Find the point in XY-plane which is equidistant from three points A(2,0,3), B(0,3,2) and C(0,0,1).

Marks:4

Ans

The z-coordinate = 0 on xy-plane.
Let P (x, y, 0) be a point on xy-plane such that PA = PB = PC
Now, PA = PB
PA2 = PB2 (x 2)2 + (y0)2 + (03)2 = (x0)2 + (y 3)2 + (02)2
4x6y = 0 or 2x 3y = 0 …(1)
PB = PC
PB2 = PC2
(x 0)2 + (y3)2 + (0 2)2 = (x 0)2 + (y0)2 + (01)2 6y + 12 = 0 or y = 2 …(2)
By (1), we get x = 3
Hence, the required point is (3, 2, 0).

Please register to view this section

FAQs (Frequently Asked Questions)

The left hand limit is the value approached when x comes from the left side of a point. The right hand limit is the value approached when x comes from the right side.

A limit exists only when both values are equal.

lim x→0 sin x / x = 1 is a standard limit used in many trigonometric limit questions.

It helps simplify expressions involving sin ax, tan ax and related trigonometric forms.

The first principle defines the derivative using a limit.

For a function f(x), f′(a) = lim h→0 [f(a + h) − f(a)] / h, if the limit exists.

A derivative shows the rate of change of a function.

In the chapter, students first understand derivative through examples such as velocity as the rate of change of distance.

Students should avoid direct substitution in 0/0 forms without simplifying. They should also check left hand and right hand limits for piecewise functions.

In trigonometric limits, angles should be treated in radians.