Class 11 Maths · Chapter 12
Short answer: <![CDATA[
Class 11 Maths Chapter 12 Limits and Derivatives ke complete NCERT solutions — algebraic limits, trigonometric limits (sin x/x, tan x/x standard results), aur derivatives (first principles + power/product/quotient rule) — sabhi CBSE marking scheme ke step-by-step format me. Ye class 11 maths limits and derivatives ncert solutions page Class 12 Calculus ki neev hai, isliye concepts clear karna especially zaroori hai.
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Chapter 12 poore Class 11 Maths Calculus ki neev hai. Yahan tak jitne chapters aaye — Sets, Relations & Functions, Trigonometric Functions, Sequences & Series, Straight Lines — sab "static" algebra tha: ek fixed value, ek fixed answer. Limits and Derivatives se pehli baar Maths me "change" aur "approach" ka idea aata hai — function kisi point ke paas kya karta hai, aur wo kitni tezi se change ho raha hai. Yehi do ideas — limit aur derivative — Class 12 ke Continuity, Application of Derivatives, aur Integrals ki poori building hain.
Is page par Class 11 Maths Chapter 12 Limits and Derivatives ke NCERT solutions step-by-step diye gaye hain — dono sections cover karte hue: pehla Limits (algebraic limits jaise (x²−a²)/(x−a) form, aur trigonometric limits jaise sin x/x), aur doosra Derivatives (first principles definition se lekar power rule, product rule aur quotient rule tak). Har solution CBSE marking scheme ke pattern me likha gaya hai — sirf final answer nahi, poora working step-by-step.
Agar tum poore Class 11 Maths NCERT solutions revise kar rahe ho, to is chapter ko Chapter 3 Trigonometric Functions ke turant baad karna best hota hai — kyunki trigonometric limits (sin x/x jaisi) trig identities pe hi based hain. Chapter 1 Sets ke basics, Chapter 8 Sequences and Series ke important questions, aur Straight Lines ke solutions bhi is hi series me available hain agar chapter-wise revision plan bana rahe ho. Neeche formulas table, common mistakes, aur exam-pattern practice questions bhi diye hain taaki last-minute revision fast ho sake.
]]>Chapter 12 Summary — 5 Minute Revision
<![CDATA[Limits and Derivatives Class 11 Maths ka Calculus-introduction chapter hai, jo do core ideas par based hai:
1. Limits — function ki value kisi point 'a' ke paas kya approach karti hai (chahe function us point pe defined ho ya na ho). Isme algebra of limits (sum, difference, product, quotient rules), polynomial/rational function limits (factor-cancel technique), aur trigonometric limits (sin x/x = 1, tan x/x = 1, jaise standard results) cover hote hain. Limit ka exist karna zaroori condition hai: Left Hand Limit (LHL) aur Right Hand Limit (RHL) dono barabar honi chahiye.
2. Derivatives — kisi function ka instantaneous rate of change, jo first principles se define hoti hai: f'(x) = limh→0 [f(x+h) − f(x)]/h. Isi definition se practical rules derive hote hain — power rule, sum/difference rule, product rule, aur quotient rule — jo polynomial aur trigonometric functions (sin x, cos x) ko differentiate karne ke liye use hote hain.
Ye poora chapter Class 12 ke Continuity & Differentiability aur Application of Derivatives ki direct foundation hai, isliye first-principles derivation samajhna sirf marks ke liye nahi, future concepts ke liye bhi zaroori hai.
]]>In-Text Questions — Solutions
Example 1. Evaluate: limx→1 (x² + 1)
Ye polynomial function hai, isliye direct substitution kaam karta hai (0/0 ya ∞/∞ form nahi hai).
limx→1 (x² + 1) = 1² + 1 = 2
Example 2. Evaluate: limx→0 (5x³ + 2x²)/x
0/0 form — pehle x common factor nikal ke cancel karo.
(5x³ + 2x²)/x = x(5x² + 2x)/x = 5x² + 2x, x ≠ 0
limx→0 (5x² + 2x) = 0 + 0 = 0
Example 3. Evaluate: limx→3 (x² − 9)/(x² − 5x + 6)
0/0 form — dono factor karo.
x² − 9 = (x − 3)(x + 3)
x² − 5x + 6 = (x − 3)(x − 2)
(x² − 9)/(x² − 5x + 6) = (x + 3)/(x − 2), x ≠ 3
limx→3 (x + 3)/(x − 2) = (3+3)/(3−2) = 6/1 = 6
Example 4. Evaluate: limx→0 sin x / x
(standard limit — derivation ka idea).
Ye chapter ka sabse important standard limit hai. Geometrically prove hota hai unit circle me sector area vs triangle area compare karke (sandwich theorem), yaha result yaad rakho:
limx→0 sin x/x = 1 (x radians me hona chahiye, degrees me nahi)
Example 5. Evaluate: limx→0 tan x / x
tan x = sin x/cos x likh ke standard limit sinx/x ka use karo.
tan x/x = (sin x/x) × (1/cos x)
limx→0 tan x/x = [lim sin x/x] × [lim 1/cos x] = 1 × (1/1) = 1
Example 6. Find the derivative of the constant function f(x) = c.
f'(x) = limh→0 [f(x+h) − f(x)]/h = limh→0 (c − c)/h = limh→0 0 = 0
Isliye kisi bhi constant function ka derivative hamesha 0 hota hai — graph ek horizontal line hai jiska slope 0 hai.
Example 7. Find the derivative of f(x) = x³ using first principles.
f'(x) = limh→0 [(x+h)³ − x³]/h
(x+h)³ = x³ + 3x²h + 3xh² + h³
f(x+h) − f(x) = 3x²h + 3xh² + h³ = h(3x² + 3xh + h²)
f'(x) = limh→0 (3x² + 3xh + h²) = 3x²
Ye power rule d/dx(xⁿ)=n·xn−1 ko confirm karta hai (n=3 case).
Example 8. Differentiate f(x) = 2x² + 3x sin x.
Do terms — pehla power rule se, dusra product rule se.
d/dx(2x²) = 4x
d/dx(3x sin x) = 3[(1)(sin x) + x(cos x)] = 3 sin x + 3x cos x
f'(x) = 4x + 3 sin x + 3x cos x
Example 9. Evaluate: limx→0 (1 − cos x)/x
(standard building-block result).
1 − cos x = 2 sin²(x/2) use karo.
(1 − cos x)/x = 2 sin²(x/2)/x = sin(x/2) × [sin(x/2)/(x/2)]
x → 0 hone par dono factors ka limit: sin(x/2) → 0, aur [sin(x/2)/(x/2)] → 1
limx→0 (1 − cos x)/x = 0 × 1 = 0
Example 10. Differentiate f(x) = x/(1 + x²) using the quotient rule.
u = x, v = 1+x², u' = 1, v' = 2x
f'(x) = [u'v − uv']/v² = [(1)(1+x²) − x(2x)]/(1+x²)²
= [1 + x² − 2x²]/(1+x²)² = (1 − x²)/(1+x²)²
f'(x) = (1 − x²)/(1 + x²)²
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q22)
Q1. Evaluate: limx→2 (x² − 4)/(x − 2)
Direct substitution x = 2 par 0/0 form deta hai, isliye factor karo.
(x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = (x + 2), x ≠ 2
∴ limx→2 (x + 2) = 2 + 2 = 4
Q2. Evaluate: limx→0 sin 3x / sin 5x
Numerator aur denominator dono ko apne coefficient se multiply-divide karo taaki standard form sinθ/θ ban jaaye.
sin 3x/sin 5x = [ (sin 3x/3x) × 3x ] / [ (sin 5x/5x) × 5x ] = (3/5) × (sin 3x/3x)/(sin 5x/5x)
x → 0 hone par 3x → 0 aur 5x → 0, isliye dono standard limit terms → 1.
limx→0 sin 3x/sin 5x = (3/5) × (1/1) = 3/5
Q3. Evaluate: limx→1 (x³ − 1)/(x − 1)
0/0 form — factor karke (x − a) cancel karo.
x³ − 1 = (x − 1)(x² + x + 1)
limx→1 (x − 1)(x² + x + 1)/(x − 1) = limx→1 (x² + x + 1)
= 1² + 1 + 1 = 3
Q4. Evaluate: limx→0 (1 − cos 2x)/x²
Identity use karo: 1 − cos 2x = 2 sin²x
(1 − cos 2x)/x² = 2 sin²x/x² = 2 (sin x/x)²
limx→0 2 (sin x/x)² = 2 × (1)² = 2
Q5. Evaluate: limx→2 (x⁵ − 32)/(x − 2)
Standard result use karo: limx→a (xⁿ − aⁿ)/(x − a) = n·an−1, yahan n = 5, a = 2.
limx→2 (x⁵ − 2⁵)/(x − 2) = 5 × 2⁴ = 5 × 16 = 80
Q6. Evaluate: limx→0 (tan x − sin x)/x³
tan x − sin x = sin x (1/cos x − 1) = sin x · (1 − cos x)/cos x
(tan x − sin x)/x³ = (sin x/x) × (1 − cos x)/x² × 1/cos x
x → 0 hone par: sin x/x → 1, (1 − cos x)/x² → 1/2 (chapter ka standard result), cos x → 1.
limx→0 (tan x − sin x)/x³ = 1 × (1/2) × 1 = 1/2
Q7. Evaluate: limx→0 (√(1+x) − 1)/x
0/0 form — rationalise karo, conjugate (√(1+x) + 1) se multiply karo.
(√(1+x) − 1)/x × (√(1+x) + 1)/(√(1+x) + 1) = [(1+x) − 1] / [x(√(1+x) + 1)]
= x/[x(√(1+x) + 1)] = 1/(√(1+x) + 1)
x → 0 substitute karo:
= 1/(1 + 1) = 1/2
Q8. Evaluate: limx→0 (cos x − 1)/x
cos x − 1 = −(1 − cos x) = −2 sin²(x/2)
(cos x − 1)/x = −2 sin²(x/2)/x = −sin(x/2) × [sin(x/2)/(x/2)]
x → 0 hone par sin(x/2) → 0 aur [sin(x/2)/(x/2)] → 1.
limx→0 (cos x − 1)/x = 0 × 1 = 0
Q9. Show that limx→0 |x|/x
does not exist.
|x|/x ki definition piecewise hai — left aur right se function alag behave karta hai, isliye LHL aur RHL alag se nikalo.
Right Hand Limit (x → 0⁺): yahan x > 0, isliye |x| = x
RHL = limx→0⁺ x/x = limx→0⁺ 1 = 1
Left Hand Limit (x → 0⁻): yahan x < 0, isliye |x| = −x
LHL = limx→0⁻ (−x)/x = limx→0⁻ (−1) = −1
Chunki LHL (−1) ≠ RHL (1), limx→0 |x|/x exist nahi karti.
Q10. Evaluate: limx→π/2 (1 − sin x)/cos x
0/0 form. cos²x = 1 − sin²x = (1 − sin x)(1 + sin x) use karo.
(1 − sin x)/cos x = (1 − sin x)/cos x × (1 + sin x)/(1 + sin x) = (1 − sin²x)/[cos x(1 + sin x)]
= cos²x/[cos x(1 + sin x)] = cos x/(1 + sin x)
x → π/2 substitute karo: cos(π/2) = 0, sin(π/2) = 1
= 0/(1 + 1) = 0
Q11. Find the derivative of f(x) = x² from first principles.
Definition:
f'(x) = limh→0 [f(x+h) − f(x)]/h
f(x+h) − f(x) = (x+h)² − x² = x² + 2xh + h² − x² = 2xh + h²
f'(x) = limh→0 (2xh + h²)/h = limh→0 (2x + h)
= 2x + 0 = 2x
Q12. Differentiate f(x) = 3x⁴ − 4x using standard rules.
Power rule d/dx(xⁿ) = n·xn−1 term by term apply karo (sum/difference rule use karte hue).
f'(x) = 3 × 4x³ − 4 × 1
f'(x) = 12x³ − 4
Q13. Differentiate f(x) = sin x from first principles.
f'(x) = limh→0 [sin(x+h) − sin x]/h
Identity use karo: sin C − sin D = 2 cos((C+D)/2) sin((C−D)/2)
sin(x+h) − sin x = 2 cos(x + h/2) sin(h/2)
f'(x) = limh→0 2 cos(x + h/2) sin(h/2)/h = limh→0 cos(x + h/2) × [sin(h/2)/(h/2)]
h → 0 hone par: cos(x + h/2) → cos x, aur [sin(h/2)/(h/2)] → 1
f'(x) = cos x × 1 = cos x
Q14. Differentiate f(x) = x sin x using the product rule.
Product rule: d/dx(uv) = u'v + uv', yahan u = x aur v = sin x.
u' = 1, v' = cos x
f'(x) = (1)(sin x) + (x)(cos x)
f'(x) = sin x + x cos x
Q15. Differentiate f(x) = (x + 1)/(x − 1) using the quotient rule.
Quotient rule: d/dx(u/v) = (u'v − uv')/v², yahan u = x+1, v = x−1
u' = 1, v' = 1
f'(x) = [(1)(x−1) − (x+1)(1)] / (x−1)²
= [x − 1 − x − 1]/(x−1)² = −2/(x−1)²
f'(x) = −2/(x − 1)²
Q16. Differentiate f(x) = cos x from first principles.
f'(x) = limh→0 [cos(x+h) − cos x]/h
Identity: cos C − cos D = −2 sin((C+D)/2) sin((C−D)/2)
cos(x+h) − cos x = −2 sin(x + h/2) sin(h/2)
f'(x) = limh→0 −2 sin(x + h/2) sin(h/2)/h = limh→0 −sin(x + h/2) × [sin(h/2)/(h/2)]
f'(x) = −sin x × 1 = −sin x
Q17. Differentiate f(x) = x³ sin x + cos x.
Do terms hain — pehla term product rule se, dusra direct standard formula se.
d/dx(x³ sin x) = 3x² sin x + x³ cos x (product rule)
d/dx(cos x) = −sin x
f'(x) = 3x² sin x + x³ cos x − sin x
Q18. Differentiate f(x) = (2x − 3)(x² + 1). Verify by two methods.
Method 1 (expand first):
f(x) = 2x³ + 2x − 3x² − 3 = 2x³ − 3x² + 2x − 3
f'(x) = 6x² − 6x + 2
Method 2 (product rule): u = 2x−3, v = x²+1, u' = 2, v' = 2x
f'(x) = u'v + uv' = 2(x² + 1) + (2x − 3)(2x)
= 2x² + 2 + 4x² − 6x = 6x² − 6x + 2
Dono method se same answer — f'(x) = 6x² − 6x + 2
Q19. Find the derivative of f(x) = 1/x from first principles.
f'(x) = limh→0 [1/(x+h) − 1/x]/h
1/(x+h) − 1/x = [x − (x+h)] / [x(x+h)] = −h/[x(x+h)]
f'(x) = limh→0 {−h/[x(x+h)]}/h = limh→0 −1/[x(x+h)]
= −1/(x · x) = −1/x²
Check: power rule se d/dx(x−1) = −1·x−2 = −1/x² — match hota hai.
Q20. Differentiate f(x) = (sin x + cos x)/(sin x − cos x).
Quotient rule: u = sin x + cos x, v = sin x − cos x
u' = cos x − sin x, v' = cos x + sin x
u'v = (cos x − sin x)(sin x − cos x) = 2 sin x cos x − 1
uv' = (sin x + cos x)(sin x + cos x) = 1 + 2 sin x cos x
u'v − uv' = (2 sin x cos x − 1) − (1 + 2 sin x cos x) = −2
v² = (sin x − cos x)² = 1 − 2 sin x cos x
f'(x) = −2/(1 − 2 sin x cos x) = −2/(1 − sin 2x)
Q21. Evaluate: limx→0 (x² cos x)/(1 − cos x)
and comment on any simplification trap.
Directly (1 − cos x) ko x²/2 se replace mat karo bina justify kiye — pehle standard limit form banao.
x²/(1 − cos x) = x²/[2 sin²(x/2)] = [x/(2 sin(x/2))]² × 2 = [ (x/2)/sin(x/2) ]² × 2
x → 0 hone par (x/2)/sin(x/2) → 1, isliye x²/(1 − cos x) → 2.
limx→0 (x² cos x)/(1 − cos x) = [lim x²/(1−cos x)] × [lim cos x] = 2 × 1 = 2
Q22. Find f'(x) if f(x) = √x, using first principles (write as x1/2 and verify with power rule).
f'(x) = limh→0 [√(x+h) − √x]/h
Rationalise: conjugate (√(x+h) + √x) se multiply karo.
= limh→0 [(x+h) − x] / [h(√(x+h) + √x)] = limh→0 h/[h(√(x+h) + √x)]
= limh→0 1/(√(x+h) + √x) = 1/(2√x)
Power rule check: d/dx(x1/2) = (1/2)x−1/2 = 1/(2√x) — match ho gaya.
Important Equations — Ek Nazar Me
<![CDATA[| Concept | Formula / Rule |
|---|---|
| Algebra of limits — sum | lim [f(x) + g(x)] = lim f(x) + lim g(x) |
| Algebra of limits — difference | lim [f(x) − g(x)] = lim f(x) − lim g(x) |
| Algebra of limits — product | lim [f(x)·g(x)] = lim f(x) × lim g(x) |
| Algebra of limits — quotient | lim [f(x)/g(x)] = lim f(x) / lim g(x), provided lim g(x) ≠ 0 |
| Algebra of limits — scalar multiple | lim [λ·f(x)] = λ × lim f(x) |
| Standard polynomial limit | limx→a (xn − an)/(x − a) = n·an−1 |
| Standard trig limit | limx→0 sin x / x = 1 (x radians me) |
| Standard trig limit | limx→0 tan x / x = 1 |
| Standard trig limit | limx→0 (1 − cos x)/x = 0 |
| Existence of limit | limx→a f(x) exists ⟺ LHL = RHL, dono finite ho |
| Derivative — first principles | f'(x) = limh→0 [f(x+h) − f(x)]/h |
| Power rule | d/dx(xn) = n·xn−1 |
| Sum/Difference rule | d/dx[f(x) ± g(x)] = f'(x) ± g'(x) |
| Product rule | d/dx[u·v] = u'v + uv' |
| Quotient rule | d/dx[u/v] = (u'v − uv')/v², v ≠ 0 |
| Derivative of sin x | d/dx(sin x) = cos x |
| Derivative of cos x | d/dx(cos x) = −sin x |
| Derivative of constant | d/dx(c) = 0 |
↔ Table ko side me swipe karein
]]>Common Mistakes — Yahan Marks Kat te Hain
- Directly limit ka answer likh dena bina check kiye ki 0/0 (indeterminate) form hai ya nahi — pehle factor/rationalise/identity apply karna zaroori hai.
- LHL aur RHL nikale bina hi limit ka answer likh dena — jaise |x|/x ya piecewise functions me limit exist hi nahi karti agar LHL ≠ RHL, ye check skip karna common mistake hai.
- Quotient rule me order galat karna — (u'v − uv')/v² ki jagah (uv' − u'v) likh dena, jisse sign ulta ho jaata hai.
- d/dx(cos x) likhte waqt negative sign bhool jaana — sahi formula −sin x hai, sin x nahi.
- sin x/x = 1 wali standard limit sirf x radians me valid hai — degrees me x use karke galat answer nikalna bahut common error hai.
- Product rule ki jagah dono functions ko independently differentiate kar dena — jaise d/dx(x³ sin x) ko galti se 3x² × cos x likh dena, jabki sahi formula u'v + uv' hai.
Board-Style Important Questions
- 1 mark: Evaluate limx→0 (sin 4x)/(sin 2x). (Answer: standard trig-limit manipulation se 2 aata hai.)
- 2 marks: Find the derivative of f(x) = x² + 1 using first principles. (First principles se hi solve karo, seedha power rule use nahi karna.)
- 3 marks: Evaluate limx→0 (tan x − sin x)/x³ — step by step trig identity ke saath solve karo.
- 3 marks: Differentiate f(x) = (x² + 1)/(x − 1) using the quotient rule and simplify fully. (Quotient rule apply karke poora simplify zaroor karo.)
- 5 marks: Find the derivative of f(x) = x³ + 1 both by first principles and by the power rule, and verify both answers match. (Dono method se same answer aana chahiye — yehi verify karna hai.)
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 12 Limits and Derivatives ke NCERT solutions me kitne exercises cover hote hain?
Chapter Limits and Derivatives (rationalised 14-chapter NCERT ke hisaab se Chapter 12) do main hisson me bata hai — pehla limits (algebraic aur trigonometric limits, direct substitution se lekar 0/0 indeterminate forms tak) aur doosra derivatives (first principles se lekar power, product, quotient rule tak). Is page ke solutions in dono sections ke representative solved questions cover karte hain, saath me kuch miscellaneous-type mixed problems bhi.
Limit aur derivative me basic difference kya hai?
Limit batati hai ki function kisi point ke paas (bina us point tak pahunche) kya value ke kareeb ja raha hai — limx→a f(x). Derivative us limit ka ek special case hai jo function ke instantaneous rate of change (slope of tangent) ko measure karta hai: f'(x) = limh→0 [f(x+h) − f(x)]/h. Simple words me — limit ek 'approach' hai, derivative us approach se banaya gaya 'rate of change' hai.
lim x→0 sin x/x ki value 1 kyu hoti hai, aur kya ye degrees me bhi 1 hoti hai?
Ye result unit circle geometry (sector area vs triangle area, sandwich/squeeze theorem) se derive hota hai, aur sirf tab valid hai jab x radians me measure ho. Agar x degrees me le liya jaaye to yahi limit π/180 ke barabar aati hai, 1 nahi — isliye exam me hamesha radians assume karna chahiye jab tak specify na ho.
Exam me first principles se derivative nikalna zaroori hai ya seedha formula use kar sakte hain?
Agar question specifically 'find derivative using first principles' bole to definition f'(x) = limh→0[f(x+h)−f(x)]/h se hi step-by-step solve karna padega, direct power/product rule use nahi kar sakte — CBSE marking scheme me steps ke liye alag marks hote hain. Agar question sirf 'differentiate' bole (first principles mention na ho) to standard rules (power, product, quotient) directly use kar sakte ho.
Class 11 Maths ke baaki chapters ke NCERT solutions aur resources kahan milenge?
Is hi site par Class 11 Maths ke saare chapters ke class 11 maths ncert solutions step-by-step available hain — jaise Chapter 1 Sets, Chapter 3 Trigonometric Functions, Chapter 8 Sequences and Series ke important questions, aur Straight Lines, Conic Sections, Probability jaise later chapters bhi. Saath me class 11 maths formulas pdf download, class 11 maths ncert book pdf download, class 11 maths syllabus 2026-27 pdf, aur class 11 maths exemplar solutions bhi separately mil jayenge revision ke liye.
Limits and Derivatives Class 12 Calculus ke liye kitna zaroori hai?
Bahut zaroori — Class 12 ke Continuity & Differentiability, Application of Derivatives, aur Integrals jaise heavy-weightage chapters is Class 11 chapter ke concepts (limit definition, first-principles derivative, power/product/quotient rule) par hi bane hain. Agar ye chapter clear nahi hai to Class 12 Calculus poora struggle ban jaata hai — isliye ise 'foundation chapter' ki tarah treat karo, sirf ek aur chapter ki tarah nahi.
Class 11 Maths — Saare Chapters
Board exam tak sirf revision karna hai?
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