Class 11 Maths NCERT Solutions

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Limits and Derivatives Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 12, Limits and Derivatives — 43 questions from Ex 12.1 to Ex 12.2, each worked through step by step in the CBSE marking pattern. Limits of polynomials, rational functions and standard forms, plus the differentiation rules and chain rule.

Class:11Subject:MathsChapter:12

4 Key Formulas

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 12?

Chapter 12 carries 2 exercise questions, numbered Ex 12.1 to Ex 12.2. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Chapter Overview

A limit asks what a function approaches as its input draws near a point; a derivative asks how fast the function is changing at that point. The chapter leans on two tools: factoring and cancelling to remove 0/0 forms, and the three standard results sinx/x → 1, (1 − cosx)/x → 0, and (eˣ − 1)/x → 1 as x → 0. Derivatives are built from first principles and then with the product, quotient and chain rules.

Board pattern

When direct substitution gives 0/0, factorise and cancel (xⁿ − aⁿ style), or rationalise surds. For the one-sided tests, compute LHL and RHL separately and compare; the limit exists only when they agree. Derivatives from first principles are a guaranteed board question — practise f′(x) = lim (f(x+h) − f(x))/h with full working.

$Ex 12.1 — Evaluating limits (32 Q)$
$Ex 12.2 — Derivatives by formula (11 Q)$
$Misc — First principles + all rules (30 Q)$

Exercise 12.1 — Evaluating Limits

32Exercise questions

Step-by-step solution

1Substitute directly: 3 + 3 = 6.

Final answer

Step-by-step solution

1Substitute directly: π − 22/7.

Final answer

π − 22/7.

Step-by-step solution

1Substitute directly: π(1)² = π.

Final answer

π.

Step-by-step solution

1Substitute directly: (4·4 + 3)/(4 − 2) = 19/2.

Final answer

19/2.

Step-by-step solution

1Substitute directly: numerator = (−1)¹⁰ + (−1)⁵ + 1 = 1 − 1 + 1 = 1.
2Denominator = −1 − 1 = −2 → limit = −1/2.

Final answer

−1/2.

Step-by-step solution

1Expand (x+1)⁵ − 1 = 5x + 10x² + 10x³ + 5x⁴ + x⁵.
2Divide by x: 5 + 10x + 10x² + 5x³ + x⁴ → as x→0: 5.

Final answer

Step-by-step solution

1Factor: 3x² − x − 10 = (3x + 5)(x − 2); x² − 4 = (x − 2)(x + 2).
2Cancel (x − 2): (3x + 5)/(x + 2) → (6 + 5)/4 = 11/4.

Final answer

11/4.

Step-by-step solution

1Factor: x⁴ − 81 = (x − 3)(x + 3)(x² + 9); 2x² − 5x − 3 = (2x + 1)(x − 3).
2Cancel (x − 3): (x + 3)(x² + 9)/(2x + 1) → (6)(18)/7 = 108/7.

Final answer

108/7.

Step-by-step solution

1Substitute x = 0: (0 + b)/(0 + 1) = b.

Final answer

Step-by-step solution

1Put y = z¹ᐟ⁶ so z¹ᐟ³ = y²; as z→1, y→1.
2(y² − 1)/(y − 1) = y + 1 → 2.

Final answer

Step-by-step solution

1Substitute directly: numerator → a + b + c and denominator → c + b + a.
2Both equal, so limit = 1.

Final answer

Step-by-step solution

1Combine: 1/x + 1/2 = (x + 2)/(2x).
2(x + 2)/(2x) ÷ (x + 2) = 1/(2x) → 1/(2·(−2)) = −1/4.

Final answer

−1/4.

Step-by-step solution

1Write as (a/b) · sin(ax)/(ax), since bx = (b/a)(ax).
2(a/b) · 1 = a/b.

Final answer

a/b.

Step-by-step solution

1Divide top and bottom by x, or use (sin ax/ax)·(bx/sin bx)·(a/b).
2= 1 · 1 · a/b = a/b.

Final answer

a/b.

Step-by-step solution

1Put y = π − x; as x→π, y→0.
2sin y/(π·y) = (1/π)(sin y/y) → (1/π)(1) = 1/π.

Final answer

1/π.

Step-by-step solution

1Substitute directly: cos 0/(π − 0) = 1/π.

Final answer

1/π.

Step-by-step solution

1cos 2x − 1 = −2sin²x and cos x − 1 = −2sin²(x/2).
2Ratio = sin²x/sin²(x/2) = (2sin(x/2)cos(x/2))²/sin²(x/2) = 4cos²(x/2).
3As x→0: 4·1 = 4.

Final answer

Step-by-step solution

1Factor x: x(a + cos x)/(b sin x).
2= (1/b)(a + cos x)·(x/sin x) → (1/b)(a + 1)·1 = (a + 1)/b.

Final answer

(a + 1)/b.

Step-by-step solution

1sec x → 1 as x→0, so x·1 → 0.

Final answer

Step-by-step solution

1Divide top and bottom by x: (a·sin(ax)/ax + b)/(a + b·sin(bx)/bx).
2= (a + b)/(a + b) = 1.

Final answer

Step-by-step solution

1cosec x − cot x = (1 − cos x)/sin x.
2= tan(x/2) → 0 (since 1 − cos x = 2sin²(x/2) and sin x = 2sin(x/2)cos(x/2)).

Final answer

Step-by-step solution

1Put y = x − π/2; as x→π/2, y→0 and tan 2x = tan(2y + π) = tan 2y.
2tan 2y/y → 2 (using tan 2y ≈ 2y near 0).

Final answer

Step-by-step solution

1At x = 0: LHL = 2(0) + 3 = 3 and RHL = 3(0 + 1) = 3 → lim(x→0) = 3.
2At x = 1: points near 1 have x > 0, so use 3(x + 1): lim(x→1) = 3(2) = 6.

Final answer

lim(x→0) = 3, lim(x→1) = 6.

Step-by-step solution

1LHL: x² − 1 → 1 − 1 = 0.
2RHL: −x² − 1 → −1 − 1 = −2.
3LHL ≠ RHL → the limit does not exist.

Final answer

Limit does not exist (LHL 0, RHL −2).

Step-by-step solution

1For x < 0, |x|/x = −1; for x > 0, |x|/x = 1.
2LHL = −1, RHL = 1 → limit does not exist.

Final answer

Does not exist (LHL −1, RHL 1).

Step-by-step solution

1For x < 0, x/|x| = −1; for x > 0, x/|x| = 1.
2LHL = −1, RHL = 1 → limit does not exist.

Final answer

Does not exist (LHL −1, RHL 1).

Step-by-step solution

1|x| is continuous everywhere, so substitute x = 5: |5| − 5 = 0.

Final answer

Step-by-step solution

1f(1) = 4 and the limit exists iff LHL = RHL = 4.
2LHL = a + b = 4; RHL = b − a = 4.
3Adding: 2b = 8 → b = 4; then a + 4 = 4 → a = 0.

Final answer

a = 0, b = 4.

Step-by-step solution

1At x = a₁ the factor (x − a₁) hits 0, so lim(x→a₁) f(x) = f(a₁) = 0.
2For a different from every aᵢ, f is continuous at a: lim(x→a) f(x) = f(a) = (a − a₁)(a − a₂)…(a − aₙ).

Final answer

lim(x→a₁) = 0; lim(x→a) = (a − a₁)(a − a₂)…(a − aₙ).

Step-by-step solution

1For a < 0, near a the branch is |x| + 1 (continuous) → limit exists.
2For a > 0, the branch is |x| − 1 (continuous) → limit exists.
3At a = 0: LHL = 1, RHL = −1 → limit does not exist.

Final answer

The limit exists for every a ≠ 0.

Step-by-step solution

1As x→1, x² − 1 → 0; for the given limit to be finite, the numerator → 0 too.
2So f(x) − 2 → 0, giving lim(x→1) f(x) = 2.

Final answer

Step-by-step solution

1At x = 0: LHL = m·0 + n = n; RHL = nx + m at 0 = m. Limit exists iff n = m.
2At x = 1: LHL = n·1 + m = n + m; RHL = n·1 + m = n + m → always equal.
3So the condition is m = n, for any integers m and n.

Final answer

m = n (any integer pair).

Exercise 12.2 — Derivatives

11Exercise questions

Step-by-step solution

1d/dx (x² − 2) = 2x.
2At x = 10: 2·10 = 20.

Final answer

20.

Step-by-step solution

1d/dx (x) = 1, so at x = 1 the derivative is 1.

Final answer

Step-by-step solution

1d/dx (99x) = 99, constant for all x.

Final answer

99.

Step-by-step solution

1(i) By the power-sum rule, d/dx(x³ − 27) = 3x².
2(ii) (x − 1)(x − 2) = x² − 3x + 2 → derivative 2x − 3.
3(iii) 1/x² = x⁻² → derivative = −2x⁻³ = −2/x³.
4(iv) Quotient rule: [(x − 1) − (x + 1)]/(x − 1)² = −2/(x − 1)².

Final answer

(i) 3x², (ii) 2x − 3, (iii) −2/x³, (iv) −2/(x − 1)².

Step-by-step solution

1Differentiate term by term: f′(x) = x⁹⁹ + x⁹⁸ + … + x + 1.
2f′(0) = 1.
3f′(1) = 1 + 1 + … + 1 (100 terms) = 100.
4Hence f′(1) = 100 = 100·f′(0). Proved.

Final answer

f′(1) = 100, f′(0) = 1 → f′(1) = 100 f′(0). Proved.

Step-by-step solution

1Differentiate each term by the power rule.
2d/dx xⁿ = nxⁿ⁻¹; d/dx (aᵏxⁿ⁻ᵏ) = (n − k)aᵏxⁿ⁻ᵏ⁻¹.

Final answer

nxⁿ⁻¹ + (n−1)axⁿ⁻² + (n−2)a²xⁿ⁻³ + … + aⁿ⁻¹.

Step-by-step solution

1(i) (x − a)(x − b) = x² − (a + b)x + ab → derivative 2x − (a + b).
2(ii) (ax² + b)² = a²x⁴ + 2abx² + b² → derivative 4a²x³ + 4abx.
3(iii) Quotient rule: [(x − b) − (x − a)]/(x − b)² = (a − b)/(x − b)².

Final answer

(i) 2x − a − b, (ii) 4a²x³ + 4abx, (iii) (a − b)/(x − b)².

Step-by-step solution

1Quotient rule with u = xⁿ − aⁿ, v = x − a: f′ = [u′v − uv′]/v².
2f′ = [nxⁿ⁻¹(x − a) − (xⁿ − aⁿ)]/(x − a)².
3Simplify the numerator: (n − 1)xⁿ − naxⁿ⁻¹ + aⁿ.

Final answer

[(n − 1)xⁿ − naxⁿ⁻¹ + aⁿ]/(x − a)².

Step-by-step solution

1(i) 2x − 3/4 → 2.
2(ii) Expand: 5x⁴ − 5x³ + 3x² − 4x + 1 → derivative 20x³ − 15x² + 6x − 4.
3(iii) x⁻³(5 + 3x) = 5x⁻³ + 3x⁻² → −15x⁻⁴ − 6x⁻³.
4(iv) x⁵(3 − 6x⁻⁹) = 3x⁵ − 6x⁻⁴ → 15x⁴ + 24x⁻⁵.
5(v) x⁻⁴(3 − 4x⁻⁵) = 3x⁻⁴ − 4x⁻⁹ → −12x⁻⁵ + 36x⁻¹⁰.
6(vi) d[2/(x+1)] = −2/(x+1)²; d[x²/(3x−1)] = (3x² − 2x)/(3x−1)². So f′ = −2/(x+1)² − (3x² − 2x)/(3x−1)².

Final answer

(i) 2, (ii) 20x³ − 15x² + 6x − 4, (iii) −15/x⁴ − 6/x³, (iv) 15x⁴ + 24/x⁵, (v) −12/x⁵ + 36/x¹⁰, (vi) −2/(x+1)² − (3x²−2x)/(3x−1)².

Step-by-step solution

1f′(x) = lim(h→0) [cos(x+h) − cos x]/h.
2cos(x+h) − cos x = −2sin(x + h/2)sin(h/2).
3= −2sin(x + h/2)·[sin(h/2)]/h → −sin x · 1 (as h→0).

Final answer

−sin x.

Step-by-step solution

1(i) sin x cos x = (1/2)sin 2x → derivative cos 2x.
2(ii) sec x → sec x tan x.
3(iii) 5 sec x + 4 cos x → 5 sec x tan x − 4 sin x.
4(iv) cosec x → −cosec x cot x.
5(v) 3 cot x + 5 cosec x → −3cosec²x − 5 cosec x cot x.
6(vi) 5 sin x − 6 cos x + 7 → 5 cos x + 6 sin x.
7(vii) 2 tan x − 7 sec x → 2sec²x − 7 sec x tan x.

Final answer

(i) cos 2x, (ii) sec x tan x, (iii) 5 sec x tan x − 4 sin x, (iv) −cosec x cot x, (v) −3cosec²x − 5 cosec x cot x, (vi) 5 cos x + 6 sin x, (vii) 2sec²x − 7 sec x tan x.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Standard limit

x \to 0 lim \frac{sin x}{x} = 1

Algebraic limit

x \to a lim \frac{x ^{2} - a ^{2}}{x - a} = 2 a

Product and quotient rule

(uv)^{'} = u^{'} v + u v^{'}, (\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}

Chain rule

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Direct substitution of indeterminate forms 0/0 always fails — factor, rationalise or use a standard limit instead of trying to evaluate directly.
Keep the product and quotient rule straight: the product rule adds two terms, the quotient rule subtracts them.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 12 (Limits and Derivatives)?

There are 2 exercise questions in this chapter, numbered Ex 12.1 to Ex 12.2. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Limits and Derivatives Class 11 Maths?

The formulas this chapter's questions actually turn on are: Standard limit, Algebraic limit, Product and quotient rule, Chain rule. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Limits and Derivatives important for JEE Main?

Very important — limits and differentiation are the gateway to the whole Class 12 calculus block, and the standard limits are asked in every board paper.

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