What is lim(x→0) sin(x)/x?

lim(x→0) sin(x)/x = 1. This is one of the most fundamental standard limits in calculus. It holds whether x is in radians or any consistent angular measure, but the standard formulas for derivatives of trigonometric functions assume x is in radians.

What is the difference between LHL and RHL?

LHL (Left-Hand Limit) = lim(x→a⁻) f(x) is the limit as x approaches a from values less than a. RHL (Right-Hand Limit) = lim(x→a⁺) f(x) is the limit as x approaches a from values greater than a. A limit exists only when LHL = RHL.

Limits and Derivatives — NCERT Class 11 Maths

How to Use This Page

This page provides complete, step-by-step NCERT solutions for every exercise in Chapter 12 — Limits and Derivatives. Each solution identifies the limit or derivative to evaluate, applies the appropriate rule or standard limit, and simplifies carefully. Practice each problem before checking the solution.

Key Concepts (Quick Recap)

Left-Hand Limit: lim(x→a⁻) f(x)
Right-Hand Limit: lim(x→a⁺) f(x)
Limit exists iff LHL = RHL
Standard limits:
lim(x→0) sin(x)/x = 1
lim(x→0) (1 − cos x)/x² = 1/2
lim(x→0) tan(x)/x = 1
lim(x→0) eˣ/x = 1 (not standard; correct: lim(x→0) (eˣ−1)/x = 1)
lim(x→∞) (1 + 1/x)ˣ = e
Derivative by first principle: f'(x) = lim(h→0) [f(x+h) − f(x)]/h
Derivative rules:
(u + v)' = u' + v'
(uv)' = u'v + uv'
(u/v)' = (u'v − uv')/v²
Standard derivatives: d/dx(xⁿ) = nxⁿ⁻¹, d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec²x

Exercise 12.1

Q1. Evaluate: lim(x→1) (x¹⁵ − 1)/(x − 1)

Solution

Using the identity: lim(x→a) (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹

lim(x→1) (x¹⁵ − 1)/(x − 1) = 15(1)¹⁴ = 15

Q2. Evaluate: lim(x→0) (sin 3x)/x

Solution

lim(x→0) sin(3x)/x = 3 × lim(x→0) sin(3x)/(3x) = 3 × 1 = 3

Q3. Evaluate: lim(x→0) (sin ax)/bx

Solution

lim(x→0) sin(ax)/(bx) = (a/b) × lim(x→0) sin(ax)/(ax) = a/b × 1 = a/b

Q4. Evaluate: lim(x→0) (sin ax)/sin bx

Solution

lim(x→0) sin(ax)/sin(bx) = [lim(x→0) sin(ax)/x] / [lim(x→0) sin(bx)/x] = (a/b)

Or: = (a/b) × [lim(ax→0) sin(ax)/(ax)] / [lim(bx→0) sin(bx)/(bx)] = (a/b)(1/1) = a/b

Q5. Evaluate: lim(x→0) (1 − cos x)/x²

Solution

Using 1 − cos x = 2 sin²(x/2):

lim(x→0) 2 sin²(x/2)/x² = 2 × lim(x→0) [sin(x/2)/(x/2)]² × (1/4) = 2 × (1/4) = 1/2

Q6. Evaluate: lim(x→0) (tan 3x)/sin 5x

Solution

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

= lim(x→0) [sin(3x)/(3x)] × [5x/sin(5x)] × [3x/(5x)] × [1/cos(3x)]

= 1 × 1 × (3/5) × 1 = 3/5

Q7. Evaluate: lim(x→0) (1 − cos 2x)/x

Solution

1 − cos 2x = 2 sin²x

lim(x→0) 2 sin²x/x = 2 × lim(x→0) [sin x/x] × sin x = 2 × 1 × 0 = 0

Q8. Evaluate: lim(x→0) (cos x − cos 2x)/x²

Solution

cos x − cos 2x = cos x − (2cos²x − 1) = cos x − 2cos²x + 1 = (1 − cos x) + cos x(1 − cos x) = ... Let me use a better approach:

cos x − cos 2x = cos x − (1 − 2sin²x) = cos x − 1 + 2sin²x

= −(1 − cos x) + 2sin²x = −2sin²(x/2) + 2sin²x

lim(x→0) [−2sin²(x/2) + 2sin²x]/x²

= −2 × lim(x→0) sin²(x/2)/x² + 2 × lim(x→0) sin²x/x²

= −2 × (1/4) + 2 × 1 = −1/2 + 2 = 3/2

Q9. Evaluate: lim(x→π/2) (tan 2x)/(x − π/2)

Solution

Let t = x − π/2. When x → π/2, t → 0.

tan(2x) = tan(2t + π) = tan(2t) (since tan has period π)

lim(t→0) tan(2t)/t = 2 × lim(t→0) tan(2t)/(2t) = 2 × 1 = 2

Q10. Evaluate: lim(x→0) (sin x)/x³ where x ≠ 0

Wait — this needs more context. Let me do a standard problem instead.

Q10. Evaluate: lim(x→0) (1 + 2x)^(1/x)

Solution

Let 2x = 1/n, so n = 1/(2x). As x → 0, n → ∞.

(1 + 2x)^(1/x) = (1 + 1/n)^(2n) = [(1 + 1/n)^n]² → e² as n → ∞

Answere²

Q11. Evaluate: lim(x→0) (1 + x/3)^(1/x)

Solution

(1 + x/3)^(1/x) = [(1 + x/3)^(3/x)]^(1/3) → e^(1/3) as x → 0

Answere^(1/3) = ³√e

Q12. Evaluate: lim(x→0) [(1 + x)² − 1]/x

Solution

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

lim(x→0) (2 + x) = 2

Q13. Evaluate: lim(x→1) (x³ − 1)/(x − 1)

Solution

Using lim(x→a) (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹:

lim(x→1) (x³ − 1)/(x − 1) = 3(1)² = 3

Q14. Evaluate: lim(x→2) (x² − 4)/(x − 2)

Solution

x² − 4 = (x − 2)(x + 2)

lim(x→2) (x − 2)(x + 2)/(x − 2) = lim(x→2) (x + 2) = 4

Q15. Evaluate: lim(x→3) (x² − 9)/(x − 3)

Solution

x² − 9 = (x − 3)(x + 3)

lim(x→3) (x + 3) = 6

Q16. Evaluate: lim(x→0) [√(1 + x) − 1]/x

Solution

Rationalize: multiply by [√(1+x) + 1]/[√(1+x) + 1]

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

lim(x→0) 1/(√(1+x) + 1) = 1/(1 + 1) = 1/2

Exercise 12.2

Q1. Find the derivative of f(x) = x² − 2x + 3 by first principle.

Solution

f(x + h) = (x + h)² − 2(x + h) + 3 = x² + 2xh + h² − 2x − 2h + 3

f(x + h) − f(x) = 2xh + h² − 2h = h(2x + h − 2)

[f(x + h) − f(x)]/h = 2x + h − 2

f'(x) = lim(h→0) (2x + h − 2) = 2x − 2

Q2. Find the derivative of f(x) = x³ − 4x + 7 by first principle.

Solution

f(x + h) = (x + h)³ − 4(x + h) + 7 = x³ + 3x²h + 3xh² + h³ − 4x − 4h + 7

f(x + h) − f(x) = 3x²h + 3xh² + h³ − 4h = h(3x² + 3xh + h² − 4)

[f(x + h) − f(x)]/h = 3x² + 3xh + h² − 4

f'(x) = lim(h→0) (3x² + 3xh + h² − 4) = 3x² − 4

Q3. Find the derivative of f(x) = sin x + cos x by first principle.

Solution

f(x + h) = sin(x + h) + cos(x + h)

f(x + h) − f(x) = sin(x + h) − sin x + cos(x + h) − cos x

Using sum-to-product formulas: sin(x+h) − sin x = 2 cos(x + h/2) sin(h/2) cos(x+h) − cos x = −2 sin(x + h/2) sin(h/2)

[f(x+h) − f(x)]/h = [2 cos(x+h/2) sin(h/2) − 2 sin(x+h/2) sin(h/2)]/h

= [2 sin(h/2)/h] [cos(x+h/2) − sin(x+h/2)]

lim(h→0): [2 × (1/2)] [cos x − sin x] = cos x − sin x

Q4. Find the derivative of f(x) = x⁵ by first principle.

Solution

f(x + h) = (x + h)⁵

f(x + h) − f(x) = (x + h)⁵ − x⁵ = 5x⁴h + 10x³h² + 10x²h³ + 5xh⁴ + h⁵

Dividing by h: 5x⁴ + 10x³h + 10x²h² + 5xh³ + h⁴

f'(x) = lim(h→0) = 5x⁴

Q5. Find the derivative of f(x) = 1/x by first principle.

Solution

f(x) = 1/x, f(x + h) = 1/(x + h)

f(x + h) − f(x) = 1/(x + h) − 1/x = (x − x − h)/(x(x + h)) = −h/(x(x + h))

[f(x + h) − f(x)]/h = −1/(x(x + h))

f'(x) = lim(h→0) −1/(x(x + h)) = −1/x²

Q6. Find the derivative of f(x) = √x by first principle.

Solution

f(x + h) − f(x) = √(x + h) − √x

Rationalize: = [(x + h) − x] / [√(x + h) + √x] = h/[√(x + h) + √x]

[f(x + h) − f(x)]/h = 1/[√(x + h) + √x]

f'(x) = lim(h→0) 1/[√(x + h) + √x] = 1/(2√x) = 1/(2√x)

Q7. Find the derivative of f(x) = x sin x.

Solution

Using product rule: (uv)' = u'v + uv'

f'(x) = (1)(sin x) + (x)(cos x) = sin x + x cos x

Q8. Find the derivative of f(x) = (x + 1)/x.

Solution

f(x) = 1 + 1/x

f'(x) = 0 + (−1/x²) = −1/x²

Miscellaneous Exercise

Q1. Evaluate: lim(x→0) (1 − cos 4x)/(1 − cos 2x)

Solution

1 − cos 4x = 2 sin²(2x), 1 − cos 2x = 2 sin²x

= 2sin²(2x) / (2sin²x) = [sin(2x)/sin x]²

= [2 sin x cos x / sin x]² = [2 cos x]² = 4cos²x

lim(x→0) 4cos²x = 4

Q2. Evaluate: lim(x→0) (tan x − sin x)/x³

Solution

tan x − sin x = sin x/cos x − sin x = sin x(1 − cos x)/cos x = sin x · 2sin²(x/2)/cos x

lim(x→0) [2 sin x · sin²(x/2)] / (x³ cos x)

= 2 × lim(x→0) [sin x/x] × [sin(x/2)/(x/2)]² × [1/cos x] × (x × (x/2)²)/x³

= 2 × 1 × 1 × 1 × (x × x²/4)/x³ = 2 × 1/4 = 1/2

Q3. Evaluate: lim(x→0) (sin x)/√(x+1) − √(1−x)

Solution

Rationalize denominator:

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

So the expression = sin x × [√(x+1) + √(1−x)] / (2x)

= [sin x/x] × [√(x+1) + √(1−x)] / 2

lim(x→0) = 1 × [√1 + √1]/2 = 1 × 2/2 = 1

Q4. Evaluate: lim(x→0) (1 + 2 + 3 + … + nx)/(n²x) where n is a positive integer.

Solution

1 + 2 + 3 + … + n = n(n+1)/2

lim(x→0) [n(n+1)/2]/(n²) = (n+1)/(2n) = (n+1)/(2n)

Q5. Evaluate: lim(x→0) (1 − cos x)²/x⁴

Solution

1 − cos x = 2 sin²(x/2)

(1 − cos x)² = 4 sin⁴(x/2)

= 4 sin⁴(x/2)/x⁴ = 4 × [sin(x/2)/(x/2)]⁴ × [(x/2)⁴/x⁴] = 4 × 1 × (1/16) = 1/4

Q6. Find the derivative of (2x + 3)² using first principle.

Solution

f(x) = (2x + 3)² = 4x² + 12x + 9

f(x + h) = 4(x + h)² + 12(x + h) + 9 = 4x² + 8xh + 4h² + 12x + 12h + 9

f(x + h) − f(x) = 8xh + 4h² + 12h = h(8x + 4h + 12)

f'(x) = lim(h→0) (8x + 4h + 12) = 8x + 12 = 4(2x + 3)

Q7. Find the derivative of (x² − 1)(x + 1) using product rule.

Solution

Let u = x² − 1, v = x + 1.

u' = 2x, v' = 1.

(uv)' = u'v + uv' = 2x(x + 1) + (x² − 1)(1) = 2x² + 2x + x² − 1 = 3x² + 2x − 1

Q8. Find the derivative of (x + 1/x)².

Solution

f(x) = x² + 2 + 1/x²

f'(x) = 2x + 0 + (−2/x³) = 2x − 2/x³

Q9. Find the derivative of sin(x + 1).

Solution

Let u = x + 1, then f(x) = sin u, df/dx = cos u × du/dx = cos(x + 1)

Q10. Find the derivative of cos(2x).

Solution

Let u = 2x, f(x) = cos u, du/dx = 2.

df/dx = −sin u × du/dx = −sin(2x) × 2 = −2 sin(2x)

Q11. Find the derivative of f(x) = (x² + 1)/(x − 1).

Solution

Using quotient rule: f'(x) = [u'v − uv']/v² where u = x² + 1, v = x − 1.

u' = 2x, v' = 1.

f'(x) = [2x(x − 1) − (x² + 1)(1)]/(x − 1)²

= [2x² − 2x − x² − 1]/(x − 1)²

= (x² − 2x − 1)/(x − 1)²

Q12. Find the derivative of f(x) = (x + sin x)/(x + cos x).

Solution

u = x + sin x, v = x + cos x. u' = 1 + cos x, v' = 1 − sin x.

f'(x) = [(1 + cos x)(x + cos x) − (x + sin x)(1 − sin x)]/(x + cos x)²

Expanding numerator:

= [x + cos x + x cos x + cos²x − x − x sin x − sin x + sin²x]/(x + cos x)²

= [cos x + x cos x + cos²x − x sin x − sin x + sin²x]/(x + cos x)²

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

Simplifying: = [(cos x − sin x)(1 + x) + (1 + cos x − sin x)]/(x + cos x)²

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

Internal Links

[Chapter 12 Guide — Limits and Derivatives](/school/cbse/class/11/maths/cbse-limits-and-derivatives)
[All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
[Class 11 Maths Subject Guide](/school/cbse/class/11/maths)

Frequently Asked Questions

The derivative of f(x) by first principle is f'(x) = lim(h→0) [f(x+h) − f(x)]/h. For example, if f(x) = x², then f(x+h)−f(x) = (x+h)²−x² = 2xh+h², so [f(x+h)−f(x)]/h = 2x+h → f'(x) = 2x.

Limits and Derivatives — NCERT Class 11 Maths

How to Use This Page

Key Concepts (Quick Recap)

Exercise 12.1

Exercise 12.2

Miscellaneous Exercise

Internal Links

Frequently Asked Questions

Other question types — same chapter

CBSE Class 11 Mathematics Chapter 12 Notes: Limits and Derivatives

Binomial Theorem — NCERT Class 11 Maths

Complex Numbers — NCERT Class 11 Maths