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Ch 12Limits and Derivativesसीमा और अवकलजLimits aur Derivatives

The idea of a limit, left and right limits, algebra of limits, limits of polynomials, rational and trigonometric functions, and derivatives from first principle and by rules — with every NCERT exercise solved.सीमा का विचार, बाएँ और दाएँ पक्ष की सीमाएँ, सीमाओं का बीजगणित, बहुपद, परिमेय और त्रिकोणमितीय फलनों की सीमाएँ, और प्रथम सिद्धांत व नियमों से अवकलज — NCERT के सभी प्रश्नों के हल के साथ।Limit ka idea, left aur right limits, limits ka algebra, polynomial, rational aur trigonometric functions ki limits, aur first principle va rules se derivatives — NCERT ke saare questions ke hal ke saath.

Didn’t understand a paragraph? Tap it (or its अ button) → हिंदी / Hinglish / Simple English.कोई पैराग्राफ़ समझ नहीं आया? उस पर (या अ बटन पर) टैप करें → English / Hinglish।Koi paragraph samajh nahi aaya? Us par (ya अ button par) tap kijiye → हिंदी / English.

👆 Words with a open a diagram or graph — tap any label to read what it means.👆 वाले शब्द पर टैप करें — चित्र या ग्राफ़ खुलेगा (हर लेबल पर टैप = उसका अर्थ)।👆 wale word par tap kijiye — diagram ya graph khulega (har label par tap = uska matlab).
📘 NCERT chapter (English PDF)📗 NCERT हिंदी PDF∑ Formulas✎ NCERT solutions🎯 Quiz

Diagrams in this chapterइस अध्याय के चित्रIs chapter ke diagrams

Derivative = slope of the tangent: the secant turns into the tangent as h → 0अवकलज = स्पर्श रेखा की ढाल: h → 0 पर छेदक रेखा स्पर्श रेखा बन जाती हैDerivative = slope of the tangent: the secant turns into the tangent as h → 0Section 12.5Left-hand and right-hand limits (Ex 12.1 Q23)बाएँ और दाएँ पक्ष की सीमाएँ (प्रश्नावली 12.1 प्रश्न 23)Left-hand and right-hand limits (Ex 12.1 Q23)Section 12.3Why lim (x→0) sin x / x = 1: comparing areas in a unit circlelim (x→0) sin x / x = 1 क्यों: इकाई वृत्त में क्षेत्रफलों की तुलनाWhy lim (x→0) sin x / x = 1: comparing areas in a unit circleSection 12.4
In this chapter (NCERT sections)इस अध्याय में (NCERT खंड)Is chapter mein (NCERT sections)
  1. 12.3LimitsसीमाएँLimits
  2. 12.4Limits of Trigonometric Functionsत्रिकोणमितीय फलनों की सीमाएँTrigonometric functions ki limits
  3. 12.5DerivativesअवकलजDerivatives
  4. ★Yaad rakho — key formulasयाद रखो — मुख्य सूत्रYaad rakho — key formulas
  5. ⚠Common mistakesआम गलतियाँCommon mistakes
  6. ✎NCERT exercises — solutionsNCERT प्रश्नावली — हलNCERT exercises — solutions
  7. 🎯Quizक्विज़Quiz

12.3LimitsसीमाएँLimits

\lim_{x \to a} f(x) = l means f(x) gets as close to l as we like when x is close enough to a (x ≠ a). Approaching from values less than a gives the left-hand limit (LHL), from values greater than a the right-hand limit (RHL). The limit exists exactly when LHL = RHL.\lim_{x \to a} f(x) = l का अर्थ है कि जब x, a के पर्याप्त पास हो (x ≠ a), तो f(x), l के जितना चाहें उतना पास आ जाता है। a से छोटे मानों से पहुँचने पर बाएँ पक्ष की सीमा (LHL), बड़े मानों से दाएँ पक्ष की सीमा (RHL)। सीमा का अस्तित्व तभी है जब LHL = RHL।\lim_{x \to a} f(x) = l ka matlab hai ki jab x, a ke kaafi paas ho (x ≠ a), to f(x), l ke jitna chahein utna paas aa jaata hai. a se chhote values se aane par left-hand limit (LHL), bade values se right-hand limit (RHL). Limit tabhi exist karti hai jab LHL = RHL.

Left-hand and right-hand limits (Ex 12.1 Q23)बाएँ और दाएँ पक्ष की सीमाएँ (प्रश्नावली 12.1 प्रश्न 23)Left-hand and right-hand limits (Ex 12.1 Q23)Section 12.3 · 4 tappable partsअनुभाग 12.3 · 4 टैप करने लायक हिस्सेSection 12.3 · 4 tappable partsOpen

12.3.1Algebra of limitsसीमाओं का बीजगणितLimits ka algebra

\lim [f(x) \pm g(x)] = \lim f(x) \pm \lim g(x)
sum / differenceयोग / अंतरjod / antar
\lim [f(x)\,g(x)] = \lim f(x) \cdot \lim g(x)
productगुणनफलproduct
\lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)},\ \lim g(x) \ne 0
quotientभागफलquotient

12.3.2Limits of polynomials and rational functionsबहुपद और परिमेय फलनों की सीमाएँPolynomial aur rational functions ki limits

For a polynomial, \lim_{x \to a} f(x) = f(a) — just substitute. For a rational function \frac{g(x)}{h(x)} with h(a) \ne 0, substitute too. If you get \frac{0}{0}, factor out (x - a), cancel, then substitute.बहुपद के लिए \lim_{x \to a} f(x) = f(a) — सीधे मान रखिए। परिमेय फलन \frac{g(x)}{h(x)} में h(a) \ne 0 हो तो भी मान रखिए। यदि \frac{0}{0} आए, तो (x - a) गुणनखंड निकालकर काटिए, फिर मान रखिए।Polynomial ke liye \lim_{x \to a} f(x) = f(a) — seedha value rakhiye. Rational function \frac{g(x)}{h(x)} mein h(a) \ne 0 ho to bhi value rakhiye. Agar \frac{0}{0} aaye, to (x - a) factor nikaal kar cancel kijiye, phir value rakhiye.

Standard limit\lim_{x \to a} \dfrac{x^n - a^n}{x - a} = n\,a^{n-1} (for any positive integer n; true for every rational n with a > 0).मानक सीमा\lim_{x \to a} \dfrac{x^n - a^n}{x - a} = n\,a^{n-1} (हर धन पूर्णांक n के लिए; a > 0 के साथ हर परिमेय n के लिए भी)।Standard limit\lim_{x \to a} \dfrac{x^n - a^n}{x - a} = n\,a^{n-1} (har positive integer n ke liye; a > 0 ke saath har rational n ke liye bhi).
ExampleEvaluate \lim_{x \to 1} \dfrac{x^{15} - 1}{x^{10} - 1}.उदाहरण\lim_{x \to 1} \dfrac{x^{15} - 1}{x^{10} - 1} का मान ज्ञात कीजिए।Example\lim_{x \to 1} \dfrac{x^{15} - 1}{x^{10} - 1} ki value nikaliye.
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. \frac{x^{15} - 1}{x - 1} \div \frac{x^{10} - 1}{x - 1} \to \frac{15 \cdot 1^{14}}{10 \cdot 1^{9}} = \frac{3}{2}

12.4Limits of Trigonometric Functionsत्रिकोणमितीय फलनों की सीमाएँTrigonometric functions ki limits

Why lim (x→0) sin x / x = 1: comparing areas in a unit circlelim (x→0) sin x / x = 1 क्यों: इकाई वृत्त में क्षेत्रफलों की तुलनाWhy lim (x→0) sin x / x = 1: comparing areas in a unit circleSection 12.4 · 7 tappable partsअनुभाग 12.4 · 7 टैप करने लायक हिस्सेSection 12.4 · 7 tappable partsOpen
\lim_{x \to 0} \frac{\sin x}{x} = 1
x in radiansx रेडियन मेंx radians mein
\lim_{x \to 0} \frac{1 - \cos x}{x} = 0
write 1 − cos x = 2 sin²(x/2)1 − cos x = 2 sin²(x/2) लिखिए1 − cos x = 2 sin²(x/2) likhiye

Trick: make the angle and the denominator match — \frac{\sin 4x}{\sin 2x} = \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \frac{4x}{2x} \to 2.तरकीब: कोण और हर को एक जैसा बनाइए — \frac{\sin 4x}{\sin 2x} = \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \frac{4x}{2x} \to 2।Trick: angle aur denominator ko same banaiye — \frac{\sin 4x}{\sin 2x} = \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \frac{4x}{2x} \to 2.

12.5DerivativesअवकलजDerivatives

Derivative = slope of the tangent: the secant turns into the tangent as h → 0अवकलज = स्पर्श रेखा की ढाल: h → 0 पर छेदक रेखा स्पर्श रेखा बन जाती हैDerivative = slope of the tangent: the secant turns into the tangent as h → 0Section 12.5 · 7 tappable partsअनुभाग 12.5 · 7 टैप करने लायक हिस्सेSection 12.5 · 7 tappable partsOpen
First principlef'(x) = \dfrac{d}{dx}f(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}, when the limit exists.प्रथम सिद्धांतf'(x) = \dfrac{d}{dx}f(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}, जब सीमा का अस्तित्व हो।First principlef'(x) = \dfrac{d}{dx}f(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}, jab limit exist kare.

12.5.1Algebra of derivativesअवकलजों का बीजगणितDerivatives ka algebra

(u \pm v)' = u' \pm v'
sum ruleयोग नियमsum rule
(uv)' = u'v + uv'
product (Leibnitz) ruleगुणनफल (लाइबनिट्ज़) नियमproduct (Leibnitz) rule
\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2},\ v \ne 0
quotient ruleभागफल नियमquotient rule

12.5.2Derivatives of polynomials and trigonometric functionsबहुपद और त्रिकोणमितीय फलनों के अवकलजPolynomial aur trigonometric functions ke derivatives

\frac{d}{dx}x^n = n x^{n-1},\quad \frac{d}{dx}(c) = 0
power ruleघात नियमpower rule
(\sin x)' = \cos x,\quad (\cos x)' = -\sin x
trigonometricत्रिकोणमितीयtrigonometric
(\tan x)' = \sec^2 x,\quad (\cot x)' = -\csc^2 x
trigonometricत्रिकोणमितीयtrigonometric
(\sec x)' = \sec x \tan x
trigonometricत्रिकोणमितीयtrigonometric
(\csc x)' = -\csc x \cot x
trigonometricत्रिकोणमितीयtrigonometric
Example 6Find the derivative of f(x) = 2x² + 3x − 5 at x = −1 and show f′(0) + 3f′(−1) = 0.उदाहरण 6f(x) = 2x² + 3x − 5 का x = −1 पर अवकलज ज्ञात कीजिए और दर्शाइए f′(0) + 3f′(−1) = 0।Example 6f(x) = 2x² + 3x − 5 ka x = −1 par derivative nikaliye aur dikhaiye f′(0) + 3f′(−1) = 0.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
  1. f'(-1) = \lim_{h \to 0} \frac{2(-1 + h)^2 + 3(-1 + h) - 5 - (-6)}{h} = \lim_{h \to 0} (2h - 1) = -1
  2. f'(0) = \lim_{h \to 0} (2h + 3) = 3 \Rightarrow 3 + 3(-1) = 0
Example 7Derivative of sin x at x = 0 from first principle.उदाहरण 7प्रथम सिद्धांत से x = 0 पर sin x का अवकलज।Example 7First principle se x = 0 par sin x ka derivative.
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. \lim_{h \to 0} \frac{\sin h - \sin 0}{h} = \lim_{h \to 0} \frac{\sin h}{h} = 1

★ Yaad rakho — key formulas★ याद रखो — मुख्य सूत्र★ Yaad rakho — key formulas

  1. Limit exists ⇔ LHL = RHL.सीमा का अस्तित्व ⇔ LHL = RHL।Limit exists ⇔ LHL = RHL.
  2. \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}
  3. \lim_{x \to 0} \frac{\sin x}{x} = 1, \lim_{x \to 0} \frac{1 - \cos x}{x} = 0, \lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b}
  4. f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
  5. (x^n)' = nx^{n-1}; (uv)' = u'v + uv'; \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}
  6. sin → cos, cos → −sin, tan → sec², sec → sec tan, cosec → −cosec cot, cot → −cosec².sin → cos, cos → −sin, tan → sec², sec → sec tan, cosec → −cosec cot, cot → −cosec²।sin → cos, cos → −sin, tan → sec², sec → sec tan, cosec → −cosec cot, cot → −cosec².

⚠ Common mistakes⚠ आम गलतियाँ⚠ Common mistakes

  • Substituting straight into 0/0 and writing "limit = 0/0" or "not defined" — factor and cancel first.0/0 में सीधे मान रखकर "सीमा = 0/0" या "परिभाषित नहीं" लिखना — पहले गुणनखंड करके काटिए।0/0 mein seedha value rakhkar "limit = 0/0" ya "defined nahi" likhna — pehle factor karke cancel kijiye.
  • Using degrees in sin x / x → 1 — the result needs radians.sin x / x → 1 में डिग्री लेना — यह परिणाम रेडियन में है।sin x / x → 1 mein degree lena — yeh result radians mein hai.
  • (uv)' = u'v' — wrong; it is u'v + uv'.(uv)' = u'v' — गलत; सही u'v + uv' है।(uv)' = u'v' — galat; sahi u'v + uv' hai.
  • Quotient rule sign order: u'v - uv' (top derivative first), not uv' - u'v.भागफल नियम में क्रम: u'v - uv' (ऊपर वाले का अवकलज पहले), uv' - u'v नहीं।Quotient rule mein order: u'v - uv' (upar wale ka derivative pehle), uv' - u'v nahi.
  • (\cos x)' = \sin x — the minus sign is missing: -\sin x.(\cos x)' = \sin x — ऋण चिह्न छूट गया: -\sin x।(\cos x)' = \sin x — minus sign chhoot gaya: -\sin x.
  • For piecewise functions, checking only one side of the point.खंडशः फलनों में बिंदु के केवल एक पक्ष को देखना।Piecewise functions mein point ke sirf ek side ko dekhna.

✎ NCERT exercises — solutions✎ NCERT प्रश्नावली — हल✎ NCERT exercises — solutions

All 73 questions of the NCERT exercises. Tap a question to see the solution — try it yourself first!NCERT प्रश्नावली के सभी 73 प्रश्न। हल देखने के लिए प्रश्न पर टैप करें — पहले स्वयं हल करके देखें!NCERT exercises ke saare 73 questions. Hal dekhne ke liye question par tap kijiye — pehle khud try kijiye!

Exercise 12.1प्रश्नावली 12.1Exercise 12.1 · 32 Q

Q1–22: evaluate the limits.प्रश्न 1–22: सीमाओं के मान ज्ञात कीजिए।Q1–22: limits ki value nikaliye.

Q1. \lim_{x \to 3} (x + 3)
  1. = 3 + 3 = 6
Q2. \lim_{x \to \pi} \left(x - \frac{22}{7}\right)
  1. = \pi - \frac{22}{7}
Q3. \lim_{r \to 1} \pi r^2
  1. = \pi(1)^2 = \pi
Q4. \lim_{x \to 4} \frac{4x + 3}{x - 2}
  1. = \frac{16 + 3}{4 - 2} = \frac{19}{2}
Q5. \lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1}
  1. = \frac{1 - 1 + 1}{-2} = -\frac{1}{2}
Q6. \lim_{x \to 0} \frac{(x + 1)^5 - 1}{x}
  1. y = x + 1 \to 1:\ \lim_{y \to 1} \frac{y^5 - 1^5}{y - 1} = 5 \cdot 1^4 = 5
Q7. \lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}
  1. \frac{(x - 2)(3x + 5)}{(x - 2)(x + 2)} \to \frac{11}{4}
Q8. \lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3}
  1. \frac{(x - 3)(x + 3)(x^2 + 9)}{(x - 3)(2x + 1)} \to \frac{6 \cdot 18}{7} = \frac{108}{7}
Q9. \lim_{x \to 0} \frac{ax + b}{cx + 1}
  1. = \frac{0 + b}{0 + 1} = b
Q10. \lim_{z \to 1} \frac{z^{1/3} - 1}{z^{1/6} - 1}
  1. \frac{z^{1/3} - 1}{z - 1} \div \frac{z^{1/6} - 1}{z - 1} \to \frac{1/3}{1/6} = 2
Q11. \lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a}, a + b + c \ne 0
  1. = \frac{a + b + c}{c + b + a} = 1
Q12. \lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}
  1. \frac{x + 2}{2x(x + 2)} = \frac{1}{2x} \to -\frac{1}{4}
Q13. \lim_{x \to 0} \frac{\sin ax}{bx}
  1. \frac{\sin ax}{ax} \cdot \frac{a}{b} \to \frac{a}{b}
Q14. \lim_{x \to 0} \frac{\sin ax}{\sin bx}, a, b \ne 0
  1. \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{a}{b} \to \frac{a}{b}
Q15. \lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)}
  1. y = \pi - x \to 0:\ \frac{1}{\pi} \cdot \frac{\sin y}{y} \to \frac{1}{\pi}
Q16. \lim_{x \to 0} \frac{\cos x}{\pi - x}
  1. = \frac{\cos 0}{\pi} = \frac{1}{\pi}
Q17. \lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}
  1. \frac{-2\sin^2 x}{-2\sin^2 \frac{x}{2}} = \frac{\left(\frac{\sin x}{x}\right)^2 x^2}{\left(\frac{\sin (x/2)}{x/2}\right)^2 \frac{x^2}{4}} \to 4
Q18. \lim_{x \to 0} \frac{ax + x\cos x}{b\sin x}
  1. \frac{x}{\sin x} \cdot \frac{a + \cos x}{b} \to \frac{a + 1}{b}
Q19. \lim_{x \to 0} x\sec x
  1. = \frac{0}{\cos 0} = 0
Q20. \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}, a, b, a + b \ne 0
  1. \div x:\ \frac{a\frac{\sin ax}{ax} + b}{a + b\frac{\sin bx}{bx}} \to \frac{a + b}{a + b} = 1
Q21. \lim_{x \to 0} (\csc x - \cot x)
  1. \frac{1 - \cos x}{\sin x} = \frac{2\sin^2 \frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \tan\frac{x}{2} \to 0
Q22. \lim_{x \to \pi/2} \frac{\tan 2x}{x - \frac{\pi}{2}}
  1. y = x - \frac{\pi}{2} \to 0:\ \tan(\pi + 2y) = \tan 2y
  2. \frac{\tan 2y}{y} = \frac{\sin 2y}{2y} \cdot \frac{2}{\cos 2y} \to 2
Q23. f(x) = 2x + 3 (x ≤ 0), 3(x + 1) (x > 0). Find \lim_{x \to 0} f(x) and \lim_{x \to 1} f(x).f(x) = 2x + 3 (x ≤ 0), 3(x + 1) (x > 0)। \lim_{x \to 0} f(x) और \lim_{x \to 1} f(x) ज्ञात कीजिए।f(x) = 2x + 3 (x ≤ 0), 3(x + 1) (x > 0). \lim_{x \to 0} f(x) aur \lim_{x \to 1} f(x) nikaliye.
  1. x \to 0:\ LHL = 3,\ RHL = 3(0 + 1) = 3 \Rightarrow 3
  2. x \to 1:\ 3(1 + 1) = 6

✔ 3 and 63 और 63 aur 6

Q24. f(x) = x² − 1 (x ≤ 1), −x² − 1 (x > 1). Find \lim_{x \to 1} f(x).f(x) = x² − 1 (x ≤ 1), −x² − 1 (x > 1)। \lim_{x \to 1} f(x) ज्ञात कीजिए।f(x) = x² − 1 (x ≤ 1), −x² − 1 (x > 1). \lim_{x \to 1} f(x) nikaliye.
  1. LHL = 1 - 1 = 0,\quad RHL = -1 - 1 = -2

✔ LHL ≠ RHL → the limit does not exist.LHL ≠ RHL → सीमा का अस्तित्व नहीं।LHL ≠ RHL → limit exist nahi karti.

Q25. f(x) = |x|/x (x ≠ 0), 0 (x = 0). Evaluate \lim_{x \to 0} f(x).f(x) = |x|/x (x ≠ 0), 0 (x = 0)। \lim_{x \to 0} f(x) ज्ञात कीजिए।f(x) = |x|/x (x ≠ 0), 0 (x = 0). \lim_{x \to 0} f(x) nikaliye.
  1. x < 0:\ \frac{-x}{x} = -1;\quad x > 0:\ \frac{x}{x} = 1

✔ LHL = −1 ≠ RHL = 1 → does not exist.LHL = −1 ≠ RHL = 1 → अस्तित्व नहीं।LHL = −1 ≠ RHL = 1 → exist nahi karti.

Q26. f(x) = x/|x| (x ≠ 0), 0 (x = 0). Find \lim_{x \to 0} f(x).f(x) = x/|x| (x ≠ 0), 0 (x = 0)। \lim_{x \to 0} f(x) ज्ञात कीजिए।f(x) = x/|x| (x ≠ 0), 0 (x = 0). \lim_{x \to 0} f(x) nikaliye.
  1. LHL = -1,\quad RHL = 1

✔ Does not exist.अस्तित्व नहीं।Exist nahi karti.

Q27. f(x) = |x| − 5. Find \lim_{x \to 5} f(x).f(x) = |x| − 5। \lim_{x \to 5} f(x) ज्ञात कीजिए।f(x) = |x| − 5. \lim_{x \to 5} f(x) nikaliye.
  1. \text{near } 5:\ |x| = x \Rightarrow 5 - 5 = 05 \text{ के पास}:\ |x| = x \Rightarrow 5 - 5 = 05 \text{ ke paas}:\ |x| = x \Rightarrow 5 - 5 = 0
Q28. f(x) = a + bx (x < 1), 4 (x = 1), b − ax (x > 1), and \lim_{x \to 1} f(x) = f(1). Find a and b.f(x) = a + bx (x < 1), 4 (x = 1), b − ax (x > 1), और \lim_{x \to 1} f(x) = f(1)। a और b ज्ञात कीजिए।f(x) = a + bx (x < 1), 4 (x = 1), b − ax (x > 1), aur \lim_{x \to 1} f(x) = f(1). a aur b nikaliye.
  1. a + b = 4,\quad b - a = 4 \Rightarrow a = 0,\ b = 4
Q29. f(x) = (x − a₁)(x − a₂)…(x − aₙ). Find \lim_{x \to a_1} f(x) and, for a ≠ a₁, …, aₙ, \lim_{x \to a} f(x).f(x) = (x − a₁)(x − a₂)…(x − aₙ)। \lim_{x \to a_1} f(x) और a ≠ a₁, …, aₙ के लिए \lim_{x \to a} f(x)।f(x) = (x − a₁)(x − a₂)…(x − aₙ). \lim_{x \to a_1} f(x) aur a ≠ a₁, …, aₙ ke liye \lim_{x \to a} f(x).
  1. \lim_{x \to a_1} f(x) = 0,\qquad \lim_{x \to a} f(x) = (a - a_1)(a - a_2)\cdots(a - a_n)
Q30. f(x) = |x| + 1 (x < 0), 0 (x = 0), |x| − 1 (x > 0). For which a does \lim_{x \to a} f(x) exist?f(x) = |x| + 1 (x < 0), 0 (x = 0), |x| − 1 (x > 0)। किन a के लिए \lim_{x \to a} f(x) का अस्तित्व है?f(x) = |x| + 1 (x < 0), 0 (x = 0), |x| − 1 (x > 0). Kin a ke liye \lim_{x \to a} f(x) exist karti hai?
  1. x \to 0:\ LHL = 1,\ RHL = -1

✔ For every a ≠ 0 (each piece is a polynomial there); not at 0.हर a ≠ 0 के लिए (वहाँ हर भाग बहुपद है); 0 पर नहीं।Har a ≠ 0 ke liye (wahan har piece polynomial hai); 0 par nahi.

Q31. If \lim_{x \to 1} \frac{f(x) - 2}{x^2 - 1} = \pi, find \lim_{x \to 1} f(x).यदि \lim_{x \to 1} \frac{f(x) - 2}{x^2 - 1} = \pi, तो \lim_{x \to 1} f(x) ज्ञात कीजिए।Agar \lim_{x \to 1} \frac{f(x) - 2}{x^2 - 1} = \pi, to \lim_{x \to 1} f(x) nikaliye.
  1. \lim (f(x) - 2) = \lim \frac{f(x) - 2}{x^2 - 1} \cdot \lim (x^2 - 1) = \pi \cdot 0 = 0

✔ \lim_{x o 1} f(x) = 2

Q32. f(x) = mx² + n (x < 0), nx + m (0 ≤ x ≤ 1), nx³ + m (x > 1). For which integers m, n do both \lim_{x \to 0} and \lim_{x \to 1} exist?f(x) = mx² + n (x < 0), nx + m (0 ≤ x ≤ 1), nx³ + m (x > 1)। किन पूर्णांकों m, n के लिए \lim_{x \to 0} और \lim_{x \to 1} दोनों का अस्तित्व है?f(x) = mx² + n (x < 0), nx + m (0 ≤ x ≤ 1), nx³ + m (x > 1). Kin integers m, n ke liye \lim_{x \to 0} aur \lim_{x \to 1} dono exist karti hain?
  1. x \to 0:\ LHL = n,\ RHL = m \Rightarrow m = n
  2. x \to 1:\ LHL = n + m = RHL \ \checkmark

✔ Limit at 0 needs m = n; limit at 1 exists for all integers m, n.0 पर सीमा के लिए m = n; 1 पर सीमा सभी पूर्णांकों m, n के लिए।0 par limit ke liye m = n; 1 par limit saare integers m, n ke liye.

Exercise 12.2प्रश्नावली 12.2Exercise 12.2 · 11 Q

Q1. Find the derivative of x² − 2 at x = 10.x = 10 पर x² − 2 का अवकलज ज्ञात कीजिए।x = 10 par x² − 2 ka derivative nikaliye.
  1. \lim_{h \to 0} \frac{(10 + h)^2 - 2 - 98}{h} = \lim_{h \to 0} (20 + h) = 20
Q2. Derivative of x at x = 1.x = 1 पर x का अवकलज।x = 1 par x ka derivative.
  1. \lim_{h \to 0} \frac{(1 + h) - 1}{h} = 1
Q3. Derivative of 99x at x = 100.x = 100 पर 99x का अवकलज।x = 100 par 99x ka derivative.
  1. \lim_{h \to 0} \frac{99(100 + h) - 9900}{h} = 99
Q4. From first principle: (i) x³ − 27 (ii) (x − 1)(x − 2) (iii) 1/x² (iv) (x + 1)/(x − 1)प्रथम सिद्धांत से: (i) x³ − 27 (ii) (x − 1)(x − 2) (iii) 1/x² (iv) (x + 1)/(x − 1)First principle se: (i) x³ − 27 (ii) (x − 1)(x − 2) (iii) 1/x² (iv) (x + 1)/(x − 1)
  1. (i)\ \lim_{h \to 0} \frac{(x + h)^3 - x^3}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2
  2. (ii)\ f(x) = x^2 - 3x + 2:\ \lim_{h \to 0} \frac{2xh + h^2 - 3h}{h} = 2x - 3
  3. (iii)\ \lim_{h \to 0} \frac{1}{h}\left[\frac{1}{(x + h)^2} - \frac{1}{x^2}\right] = \lim_{h \to 0} \frac{-(2x + h)}{x^2(x + h)^2} = -\frac{2}{x^3}
  4. (iv)\ \lim_{h \to 0} \frac{1}{h}\left[\frac{x + h + 1}{x + h - 1} - \frac{x + 1}{x - 1}\right] = \lim_{h \to 0} \frac{-2}{(x + h - 1)(x - 1)} = -\frac{2}{(x - 1)^2}
Q5. f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1. Prove f′(1) = 100 f′(0).f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1। सिद्ध कीजिए f′(1) = 100 f′(0)।f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1. Prove kijiye f′(1) = 100 f′(0).
  1. f'(x) = x^{99} + x^{98} + \dots + x + 1
  2. f'(1) = 100,\quad f'(0) = 1 \Rightarrow f'(1) = 100 f'(0)
Q6. Derivative of x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n.x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n का अवकलज।x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n ka derivative.
  1. nx^{n-1} + a(n - 1)x^{n-2} + a^2(n - 2)x^{n-3} + \dots + a^{n-1}
Q7. (i) (x − a)(x − b) (ii) (ax² + b)² (iii) (x − a)/(x − b)
  1. (i)\ (x - a) + (x - b) = 2x - a - b
  2. (ii)\ 2(ax^2 + b) \cdot 2ax = 4ax(ax^2 + b)
  3. (iii)\ \frac{(x - b) - (x - a)}{(x - b)^2} = \frac{a - b}{(x - b)^2}
Q8. Derivative of \frac{x^n - a^n}{x - a}.\frac{x^n - a^n}{x - a} का अवकलज।\frac{x^n - a^n}{x - a} ka derivative.
  1. \frac{nx^{n-1}(x - a) - (x^n - a^n)}{(x - a)^2} = \frac{nx^n - anx^{n-1} - x^n + a^n}{(x - a)^2}
Q9. (i) 2x − 3/4 (ii) (5x³ + 3x − 1)(x − 1) (iii) x⁻³(5 + 3x) (iv) x⁵(3 − 6x⁻⁹) (v) x⁻⁴(3 − 4x⁻⁵) (vi) 2/(x + 1) − x²/(3x − 1)
  1. (i)\ 2\qquad (ii)\ (15x^2 + 3)(x - 1) + (5x^3 + 3x - 1) = 20x^3 - 15x^2 + 6x - 4
  2. (iii)\ 5x^{-3} + 3x^{-2} \to -15x^{-4} - 6x^{-3} = -\frac{3(5 + 2x)}{x^4}
  3. (iv)\ 3x^5 - 6x^{-4} \to 15x^4 + \frac{24}{x^5}\qquad (v)\ 3x^{-4} - 4x^{-9} \to -\frac{12}{x^5} + \frac{36}{x^{10}}
  4. (vi)\ -\frac{2}{(x + 1)^2} - \frac{2x(3x - 1) - 3x^2}{(3x - 1)^2} = -\frac{2}{(x + 1)^2} - \frac{x(3x - 2)}{(3x - 1)^2}
Q10. Derivative of cos x from first principle.प्रथम सिद्धांत से cos x का अवकलज।First principle se cos x ka derivative.
  1. \lim_{h \to 0} \frac{\cos(x + h) - \cos x}{h} = \lim_{h \to 0} \frac{-2\sin\left(x + \frac{h}{2}\right)\sin\frac{h}{2}}{h}
  2. = -\lim_{h \to 0} \sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h/2} = -\sin x
Q11. (i) sin x cos x (ii) sec x (iii) 5 sec x + 4 cos x (iv) cosec x (v) 3 cot x + 5 cosec x (vi) 5 sin x − 6 cos x + 7 (vii) 2 tan x − 7 sec x
  1. (i)\ \cos^2 x - \sin^2 x = \cos 2x\qquad (ii)\ \sec x \tan x
  2. (iii)\ 5\sec x\tan x - 4\sin x\qquad (iv)\ -\csc x \cot x
  3. (v)\ -3\csc^2 x - 5\csc x\cot x\qquad (vi)\ 5\cos x + 6\sin x
  4. (vii)\ 2\sec^2 x - 7\sec x\tan x

Miscellaneous Exercise on Chapter 12अध्याय 12 पर विविध प्रश्नावलीMiscellaneous Exercise on Chapter 12 · 30 Q

Q2–30: find the derivative (a, b, c, d, p, q, r, s are fixed non-zero constants; m, n integers).प्रश्न 2–30: अवकलज ज्ञात कीजिए (a, b, c, d, p, q, r, s अचर अशून्य; m, n पूर्णांक)।Q2–30: derivative nikaliye (a, b, c, d, p, q, r, s fixed non-zero constants; m, n integers).

Q1. From first principle: (i) −x (ii) (−x)⁻¹ (iii) sin(x + 1) (iv) cos(x − π/8)प्रथम सिद्धांत से: (i) −x (ii) (−x)⁻¹ (iii) sin(x + 1) (iv) cos(x − π/8)First principle se: (i) −x (ii) (−x)⁻¹ (iii) sin(x + 1) (iv) cos(x − π/8)
  1. (i)\ \lim_{h \to 0} \frac{-(x + h) + x}{h} = -1
  2. (ii)\ -\frac{1}{x}:\ \lim_{h \to 0} \frac{1}{h}\left[-\frac{1}{x + h} + \frac{1}{x}\right] = \lim_{h \to 0} \frac{1}{x(x + h)} = \frac{1}{x^2}
  3. (iii)\ \lim_{h \to 0} \frac{2\cos\left(x + 1 + \frac{h}{2}\right)\sin\frac{h}{2}}{h} = \cos(x + 1)
  4. (iv)\ \lim_{h \to 0} \frac{-2\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right)\sin\frac{h}{2}}{h} = -\sin\left(x - \frac{\pi}{8}\right)
Q2. (x + a)
  1. 1
Q3. (px + q)\left(\frac{r}{x} + s\right)
  1. = pr + psx + \frac{qr}{x} + qs \to ps - \frac{qr}{x^2}
Q4. (ax + b)(cx + d)²
  1. a(cx + d)^2 + 2c(ax + b)(cx + d)
Q5. \frac{ax + b}{cx + d}
  1. \frac{a(cx + d) - c(ax + b)}{(cx + d)^2} = \frac{ad - bc}{(cx + d)^2}
Q6. \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}, x ≠ 0
  1. = \frac{x + 1}{x - 1} \to \frac{(x - 1) - (x + 1)}{(x - 1)^2} = \frac{-2}{(x - 1)^2},\ x \ne 0, 1
Q7. \frac{1}{ax^2 + bx + c}
  1. -\frac{2ax + b}{(ax^2 + bx + c)^2}
Q8. \frac{ax + b}{px^2 + qx + r}
  1. \frac{a(px^2 + qx + r) - (ax + b)(2px + q)}{(px^2 + qx + r)^2} = \frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2}
Q9. \frac{px^2 + qx + r}{ax + b}
  1. \frac{(2px + q)(ax + b) - a(px^2 + qx + r)}{(ax + b)^2} = \frac{apx^2 + 2bpx + bq - ar}{(ax + b)^2}
Q10. \frac{a}{x^4} - \frac{b}{x^2} + \cos x
  1. -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x
Q11. 4\sqrt{x} - 2
  1. 4 \cdot \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x}}
Q12. (ax + b)ⁿ
  1. na(ax + b)^{n-1}
Q13. (ax + b)ⁿ(cx + d)ᵐ
  1. na(ax + b)^{n-1}(cx + d)^m + mc(ax + b)^n(cx + d)^{m-1}
  2. = (ax + b)^{n-1}(cx + d)^{m-1}\left[mc(ax + b) + na(cx + d)\right]
Q14. sin(x + a)
  1. \cos(x + a)
Q15. cosec x cot x
  1. (-\csc x\cot x)\cot x + \csc x(-\csc^2 x) = -\csc x\cot^2 x - \csc^3 x
Q16. \frac{\cos x}{1 + \sin x}
  1. \frac{-\sin x(1 + \sin x) - \cos^2 x}{(1 + \sin x)^2} = \frac{-(1 + \sin x)}{(1 + \sin x)^2} = \frac{-1}{1 + \sin x}
Q17. \frac{\sin x + \cos x}{\sin x - \cos x}
  1. \frac{-(\sin x - \cos x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2} = \frac{-2}{(\sin x - \cos x)^2}
Q18. \frac{\sec x - 1}{\sec x + 1}
  1. \frac{\sec x\tan x\,[(\sec x + 1) - (\sec x - 1)]}{(\sec x + 1)^2} = \frac{2\sec x\tan x}{(\sec x + 1)^2}
Q19. sinⁿ x
  1. n\sin^{n-1}x\cos x
Q20. \frac{a + b\sin x}{c + d\cos x}
  1. \frac{b\cos x(c + d\cos x) + d\sin x(a + b\sin x)}{(c + d\cos x)^2} = \frac{bc\cos x + ad\sin x + bd}{(c + d\cos x)^2}
Q21. \frac{\sin(x + a)}{\cos x}
  1. \frac{\cos(x + a)\cos x + \sin(x + a)\sin x}{\cos^2 x} = \frac{\cos a}{\cos^2 x}
Q22. x⁴(5 sin x − 3 cos x)
  1. 4x^3(5\sin x - 3\cos x) + x^4(5\cos x + 3\sin x)
  2. = x^3(5x\cos x + 3x\sin x + 20\sin x - 12\cos x)
Q23. (x² + 1) cos x
  1. 2x\cos x - (x^2 + 1)\sin x
Q24. (ax² + sin x)(p + q cos x)
  1. (2ax + \cos x)(p + q\cos x) - q\sin x\,(ax^2 + \sin x)
Q25. (x + cos x)(x − tan x)
  1. (1 - \sin x)(x - \tan x) + (x + \cos x)(1 - \sec^2 x)
  2. = (1 - \sin x)(x - \tan x) - \tan^2 x\,(x + \cos x)
Q26. \frac{4x + 5\sin x}{3x + 7\cos x}
  1. \frac{(4 + 5\cos x)(3x + 7\cos x) - (4x + 5\sin x)(3 - 7\sin x)}{(3x + 7\cos x)^2}
  2. = \frac{35 + 15x\cos x + 28\cos x + 28x\sin x - 15\sin x}{(3x + 7\cos x)^2}
Q27. \frac{x^2\cos\frac{\pi}{4}}{\sin x}
  1. \cos\frac{\pi}{4} \cdot \frac{2x\sin x - x^2\cos x}{\sin^2 x} = \frac{x\cos\frac{\pi}{4}\,(2\sin x - x\cos x)}{\sin^2 x}
Q28. \frac{x}{1 + \tan x}
  1. \frac{(1 + \tan x) - x\sec^2 x}{(1 + \tan x)^2}
Q29. (x + sec x)(x − tan x)
  1. (1 + \sec x\tan x)(x - \tan x) + (x + \sec x)(1 - \sec^2 x)
Q30. \frac{x}{\sin^n x}
  1. \frac{\sin^n x - nx\sin^{n-1}x\cos x}{\sin^{2n}x} = \frac{\sin x - nx\cos x}{\sin^{n+1}x}

🎯 Quiz — 13 questions🎯 क्विज़ — 13 प्रश्न🎯 Quiz — 13 questions

Tap an option. The answer and a short reason appear at once.किसी विकल्प पर टैप करें। उत्तर और छोटा कारण तुरंत दिखेगा।Kisi option par tap kijiye. Answer aur chhota reason turant dikhega.

1. \lim_{x \to 2} (x^2 + 3x) equals\lim_{x \to 2} (x^2 + 3x) बराबर है\lim_{x \to 2} (x^2 + 3x) barabar hai

2. \lim_{x \to 1} \frac{x^2 - 1}{x - 1} equals\lim_{x \to 1} \frac{x^2 - 1}{x - 1} बराबर है\lim_{x \to 1} \frac{x^2 - 1}{x - 1} barabar hai

3. \lim_{x \to 0} \frac{\sin 5x}{x} equals\lim_{x \to 0} \frac{\sin 5x}{x} बराबर है\lim_{x \to 0} \frac{\sin 5x}{x} barabar hai

4. \lim_{x \to 0} \frac{1 - \cos x}{x} equals\lim_{x \to 0} \frac{1 - \cos x}{x} बराबर है\lim_{x \to 0} \frac{1 - \cos x}{x} barabar hai

5. \lim_{x \to 2} \frac{x^5 - 32}{x - 2} equals\lim_{x \to 2} \frac{x^5 - 32}{x - 2} बराबर है\lim_{x \to 2} \frac{x^5 - 32}{x - 2} barabar hai

6. lim (x→a) f(x) exists exactly whenlim (x→a) f(x) का अस्तित्व तभी है जबlim (x→a) f(x) tabhi exist karti hai jab

7. The derivative of x⁷ isx⁷ का अवकलज हैx⁷ ka derivative hai

8. d/dx (cos x) =

9. d/dx (x sin x) =

10. d/dx (1/x) =

11. If f(x) = 3x² + 2x, then f′(1) =यदि f(x) = 3x² + 2x, तो f′(1) =Agar f(x) = 3x² + 2x, to f′(1) =

12. d/dx (tan x) =

13. The derivative f′(a) is geometrically theज्यामितीय रूप से f′(a) हैGeometrically f′(a) hai

← Ch 11Ch 13 →

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These are Simple Learning Hub's own study notes and solutions, written to follow the NCERT Class 11 Mathematics textbook (Reprint 2026-27) section by section; read them with the official NCERT chapter. Diagrams are original SLH drawings.ये Simple Learning Hub के अपने स्टडी नोट्स हैं, जो NCERT कक्षा 11 गणित की किताब (2026-27) के हर अनुभाग के साथ लिखे गए हैं; इन्हें NCERT के आधिकारिक अध्याय के साथ पढ़िए। डायग्राम SLH के अपने बनाए चित्र हैं।Ye Simple Learning Hub ke apne study notes hain, jo NCERT Class 11 Maths ki kitaab (2026-27) ke har section ke saath likhe gaye hain; inhe NCERT ke official chapter ke saath padhiye. Diagrams SLH ki apni drawings hain.