Limits & Derivatives 🚀

Calculus ka pehla step! "Approaching" ka concept — ek point ke bilkul paas jaana lekin wahan pahunchna nahi. Yahaan se shuru hota hai modern mathematics.

🚗 Car ki Speed — Limit ka Intuition

Socho tumhari car ek junction ki taraf aa rahi hai — speed 60 km/h se 50, 40, 30 hoti ja rahi hai. Junction par pahunchoge toh ruk jaogi — speed 0. Lekin just before junction par speed kitni thi? Agar hum bahut chhotay time intervals mein speed measure karein — 0.1 sec, 0.01 sec, 0.001 sec pehle — toh ek specific value par approach karte hain. Yahi hai Limit ka concept! Aur agar woh speed measure kar rahe ho — change in position per change in time — yahi hai Derivative!

1. Limit kya hai? — Intuition

Limit batata hai ki jab x kisi value 'a' ke paas bahut paas aata hai (bilkul equal nahi, sirf approach karta hai) toh f(x) kis value ke paas aata hai.

lim(x→a) f(x) = L

Padhte hain: "x approaches a ke liye f(x) ki limit L hai". Important: x = a par f(x) defined hona zaroori nahi — bas approach karna zaroori hai!

Do sides se approach hota hai: Left Hand Limit (x→a⁻) aur Right Hand Limit (x→a⁺). Limit exist karta hai sirf tab jab LHL = RHL.

f(x) ka limit x→2 par — x=2 par hole (not defined) lekin limit exist karta hai

x→2⁻ x→2⁺ Limit=4 x=2 f(x)

2. Limit Nikalne ke Tareeqe

① Direct Substitution

Seedha x=a rakh do. Agar answer 0/0 nahi aata → yahi limit hai.

lim(x→3) x²+1 = 9+1 = 10

② Factorisation

0/0 form mile toh numerator/denominator factorize karo, common factor cancel karo.

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

③ Rationalisation

Square roots ho toh conjugate se multiply karo.

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

④ Standard Results

Yeh results yaad karo — exams mein seedha use hote hain:

lim(x→0) sinx/x = 1
lim(x→0) (1−cosx)/x = 0
lim(x→a) (xⁿ−aⁿ)/(x−a) = naⁿ⁻¹

✅ lim(x→0) sin(3x)/x nikalo

sin(3x)/x = 3·sin(3x)/(3x)
= 3 · [sin(3x)/(3x)] → 3 × 1 = 3   [kyunki lim(t→0) sint/t = 1]

Trick: Argument same hona chahiye — sin(3x)/(3x) ka limit 1 hai, toh 3x/x = 3 bacha.

3. Derivative — Rate of Change

Derivative f'(x) batata hai ki x mein ek chota sa change dx hone par f(x) mein kitna change dx hota hai — yaani instant rate of change. Geometrically, yeh curve par ek point par tangent line ka slope hai.

Yeh First Principle (ya Limit Definition) se define hota hai:

f'(x) = lim(h→0) [f(x+h) − f(x)] / h

Ise "dy/dx" ya "f'(x)" ya "d/dx[f(x)]" likhte hain. Iska physical meaning: velocity = derivative of position with respect to time!

Derivative = tangent line ka slope at a point

P(x, f(x)) Tangent line slope = f'(x) y=x² f'(x) = 2x

4. First Principle se Derivative

✅ f(x) = x² ka derivative First Principle se nikalo

f'(x) = lim(h→0) [f(x+h) − f(x)] / h
= lim(h→0) [(x+h)² − x²] / h
= lim(h→0) [x²+2xh+h² − x²] / h
= lim(h→0) [2xh + h²] / h = lim(h→0) (2x + h)
h→0 rakhne par = 2x

Yahi hai Power Rule: d/dx(xⁿ) = nxⁿ⁻¹ ka basis!

5. Standard Derivative Formulas

Power Rule

d/dx(xⁿ) = nxⁿ⁻¹

Constant

d/dx(c) = 0

sin x

d/dx(sinx) = cosx

cos x

d/dx(cosx) = −sinx

tan x

d/dx(tanx) = sec²x

eˣ

d/dx(eˣ) = eˣ

ln x

d/dx(lnx) = 1/x

Constant×f

d/dx(cf) = c·f'

6. Rules — Product, Quotient, Chain

Sum/Difference Rule: d/dx[f±g] = f'±g' — seedha alag alag differentiate karo.

Product Rule: d/dx[f·g] = f'g + fg'
Quotient Rule: d/dx[f/g] = (f'g − fg') / g²

Mnemonic for Product Rule: "d-first times second PLUS first times d-second"

✅ f(x) = x³ + 3x² − 5x + 2 differentiate karo

f'(x) = 3x² + 3(2x) − 5(1) + 0
= 3x² + 6x − 5

✅ Product Rule: f(x) = x² · sinx differentiate karo

f'(x) = d/dx(x²)·sinx + x²·d/dx(sinx)
= 2x·sinx + x²·cosx
= x(2sinx + x·cosx)

7. Practice Questions

Q1. lim(x→2) (x²−4)/(x−2) nikalo.

Solution

= lim(x+2)(x−2)/(x−2) = lim(x+2) = 4

Q2. lim(x→0) sin(5x)/tan(3x) nikalo.

Solution

= lim [sin(5x)/5x] × [5x/3x] × [3x/tan(3x)] = 1 × 5/3 × 1 = 5/3

Q3. f(x) = 3x⁴ − 2x³ + x − 7 differentiate karo.

Solution

f'(x) = 12x³ − 6x² + 1

Q4. First Principle se d/dx(sinx) nikalo.

Solution

= lim[sin(x+h)−sinx]/h = lim[sinx·cosh+cosx·sinh−sinx]/h = lim[sinx(cosh−1)/h + cosx·sinh/h] = sinx·0 + cosx·1 = cosx

Q5. f(x) = (x+1)/(x−1) differentiate karo using Quotient Rule.

Solution

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

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