Limit laws — sum, product, quotient, constant multiple

4.1.2 · Maths › Calculus I — Limits & Derivatives

YE LAWS KYA HAIN?

Wo do atoms jinse sab kuch banta hai: $lim_{x \to a} c = c and lim_{x \to a} x = a$

Figure — Limit laws — sum, product, quotient, constant multiple

YE SACH KYUN HAIN? (Scratch se derivation)

Ek limit $lim_{x \to a} f (x) = L$ ka matlab hai: hum $f (x)$ ko $L$ ke jitna chahein utna kareeb force kar sakte hain bas $x$ ko $a$ ke paas rakh ke. Formally, har $ε > 0$ ke liye ek $δ > 0$ hota hai aisa ki $0 < ∣ x - a ∣ < δ \Rightarrow ∣ f (x) - L ∣ < ε .$

Sum law (poora proof)

Hum chahte hain ki $∣ (f + g) - (L + M) ∣$ chhota ho. Regroup karo aur triangle inequality $∣ p + q ∣ \leq ∣ p ∣ + ∣ q ∣$ use karo: $∣ (f (x) + g (x)) - (L + M) ∣ = ∣ (f (x) - L) + (g (x) - M) ∣ \leq ∣ f (x) - L ∣ + ∣ g (x) - M ∣.$ Ye step kyun? Hum total error ko " $f$ ka error" + " $g$ ka error" mein tod dete hain, jinmein se har ek ko hum control kar sakte hain.

Diye gaye $ε > 0$ ke liye, $δ_{1}$ choose karo jo $∣ f - L ∣ < \frac{ε}{2}$ banaye aur $δ_{2}$ jo $∣ g - M ∣ < \frac{ε}{2}$ banaye. $δ = min (δ_{1}, δ_{2})$ lo. Tab dono ek saath hold hote hain: $∣ (f + g) - (L + M) ∣ < \frac{ε}{2} + \frac{ε}{2} = ε . ■$ $ε$ ko $ε /2 + ε /2$ mein kyun split kiya? Taaki dono halves milke exactly $ε$ ban jayein.

Constant multiple (proof)

$∣ c f (x) - c L ∣ = ∣ c ∣ ∣ f (x) - L ∣.$ Agar $c \neq = 0$ hai, toh $∣ f - L ∣ < \frac{ε}{∣ c ∣}$ banao, toh product $< ε$ ho jaayega. (Agar $c = 0$ toh dono sides $0$ hain.) $∣ c ∣$ se divide kyun kiya? Taaki $∣ c ∣$ se wapas multiply karne par $ε$ mile.

Product (key trick ki sketch)

Clever move hai add aur subtract $L g (x)$ : $f g - L M = f g - Lg + Lg - L M = g (f - L) + L (g - M) .$ Ye step kyun? Ye " $f$ ka wobbling" ko " $g$ ke wobbling" se alag karta hai. $a$ ke paas, $g$ bounded hai ( $M$ ke kareeb), isliye $g (f - L) \to 0$ , aur $L (g - M) \to 0$ . Dono $0$ par jaate hain, isliye product limit $L M$ hai.

Quotient

Pehle prove karo ki $lim \frac{1}{g} = \frac{1}{M}$ (iske liye $M \neq = 0$ chahiye taaki $g$ paas mein $0$ se door rahe), phir $f \cdot \frac{1}{g}$ par product law apply karo. $M \neq = 0$ kyun chahiye? $0$ ki taraf jaane wali cheez se divide karna "blow up" kar deta hai — result exist karna zaroori nahi.

INHE KAISE USE KAREIN — worked examples

Common mistake Steel-manned mistakes

Mistake A: " $lim \frac{f}{g} = \frac{L}{M}$ hamesha." Kyun sahi lagta hai: law unconditional lagta hai. Fix: iska ek fine-print clause hai $M \neq = 0$ . Agar $M = 0$ hai, toh law kuch nahi deta — tumhe simplify, rationalize karna hoga, ya one-sided/infinite behaviour check karna hoga.

Mistake B: " $lim (f + g)$ exist karta hai, toh $lim f$ aur $lim g$ dono exist karte honge." Kyun sahi lagta hai: lagta hai jaise sum law ko backwards chala sakte hain. Fix: GALAT hai. Laws maan ke chalte hain ki parts ke limits pehle exist karte hain. E.g. $f = ⌊ x ⌋, g = - ⌊ x ⌋$ kisi integer par: dono jump karte hain, lekin $f + g \equiv 0$ . Tum sirf known existing limits ko combine kar sakte ho.

Mistake C: $\frac{x - 3}{x - 3} = 1$ "at $x = 3$ " cancel karna. Kyun sahi lagta hai: algebra mein usually cancel kar sakte hain. Fix: exactly $x = 3$ par ye $\frac{0}{0}$ hai; lekin limit sirf $x$ ke $3$ ke paas hone ki parwah karta hai, jahan $x \neq = 3$ hai, isliye cancel karna limit ke liye valid hai.

Recall Feynman: 12-saal ke bacche ko samjhao

Socho do dost ek meeting spot ki taraf chal rahe hain. Ek $L$ kursi par pahunchta hai, doosra $M$ kursi par. Agar tum unke beech ki doori measure karo, toh wo $L + M$ ki taraf jaayegi; agar tum unhe stack karo, toh $L \cdot M$ . Jaanna ki har dost kahan ja raha hai tumhe batata hai ki koi bhi combination kahan ja raha hai — jab tak tum kisi aisi dost se divide karne ki koshish nahi karte jo seedha zero par chala jaata hai, kyunki tab "share per person" explode ho jaata hai. Pura game yehi hai: limits $+, -, \times$ se slide kar jaate hain, aur $\div$ se tabhi jab bottom zero ki taraf nahi ja raha ho.

Flashcards

Sum law statement

lim (f + g) = lim f + lim g = L + M

, jab dono limits exist karti hain.

Constant multiple law

lim (c f) = c lim f = c L

Product law

lim (f g) = (lim f) (lim g) = L M

Quotient law + condition

lim (f / g) = L / M

, valid tabhi jab

M \neq = 0

Do atomic limits jinse sab kuch banta hai

lim_{x \to a} c = c

aur

lim_{x \to a} x = a

Product-law proof mein key trick

Add & subtract

Lg (x)

f g - L M = g (f - L) + L (g - M)

Sum proof mein

ε

ε /2

mein split kyun kiya

Taaki do bounded errors milke exactly

ε

ban jayein.

Jab quotient

0/0

de toh kya karo

Law fail hota hai; factor/cancel/rationalize karo, phir limit lo.

Kya

lim (f + g)

ka exist karna imply karta hai ki har limit exist karti hai

Nahi — counterexample

⌊ x ⌋ + (- ⌊ x ⌋)

a

par polynomial ki limit

p (a)

— direct substitution (sum + const-mult + power laws se).

Connections

Epsilon-Delta definition of a limit — wo foundation jisse har law prove hota hai.
One-sided limits — laws bilkul usi tarah left/right limits par bhi apply hote hain.
Indeterminate forms 0 over 0 — jab quotient law fail ho tab kya karo.
Continuity — ek function $a$ par continuous hai iff $lim_{x \to a} f = f (a)$ ; limit laws ⇒ continuous functions ke sums/products continuous hote hain.
Squeeze theorem — wo tool jo un limits ke liye hai jahan algebra laws nahi pahunch sakte.
Derivative as a limit — har derivative rule (sum/product/quotient rule) inhi laws se aata hai.