Limit laws — sum, product, quotient, constant multiple: Question bank

4.1.2 · D5 · Maths › Calculus I — Limits & Derivatives › Limit laws — sum, product, quotient, constant multiple

True ya false — justify karo

Har convergent limit law ko pehle parts ke limits ka exist karna zaroori hai.

True. Statement "if

lim f = L

and

lim g = M

then..." ek implication hai; jab hypotheses fail hote hain, conclusion simply claim nahi kiya jaata — parent mein Mistake B dekho.

lim (f + g)

exist karna matlab dono

lim f

aur

lim g

exist karte hain.

False. Counterexample:

f = ⌊ x ⌋

g = - ⌊ x ⌋

kisi integer par — dono jump karte hain aur koi limit nahi hai, phir bhi

f + g \equiv 0

ka limit

0

hai.

lim (f g)

exist karna matlab

lim f

aur

lim g

dono exist karte hain.

False.

f (x) = x

aur

g (x) = sin (1/ x)

0

ke paas:

g

ka koi limit nahi, lekin

f g \to 0

Squeeze theorem se. Product converge kar sakta hai jabki ek factor nahi karta.

Agar

lim f = L

exist karta hai lekin

lim g

nahi karta, toh

lim (f + g)

exist nahi kar sakta.

True. Agar

lim (f + g)

exist karta, toh

g = (f + g) - f

ka difference law se ek limit hota — contradiction. Isliye sum,

g

ki non-existence inherit karta hai.

Quotient law

lim (f / g) = L / M

tab holds karta hai jab bhi

M \neq = 0

ho, chahe

L

kuch bhi ho.

True.

M \neq = 0

ek maatra condition hai;

L

kuch bhi ho sakta hai,

0

bhi (tab quotient limit bas

0

hai).

Agar

M = 0

toh

lim (f / g)

ka exist karna fail hona zaroori hai.

False. Agar

L = 0

bhi ho toh tumhe indeterminate

0/0

milta hai, jiska factoring ke baad finite limit ho sakta hai — jaise

\frac{x ^{2} - 9}{x - 3} \to 6

. Indeterminate forms 0 over 0 dekho.

Power law

lim [f]^{n} = L^{n}

ke liye

n

ka positive integer hona zaroori hai.

Repeated product se proof ke liye True. General real exponents ke liye tumhe

x \mapsto x^{n}

ki continuity aur ek positivity/limit condition chahiye, sirf product law nahi.

Direct substitution

lim_{x \to a} p (x) = p (a)

har function ke liye kaam karta hai, sirf polynomials ke liye nahi.

False. Ye tab kaam karta hai jab function

a

par continuous ho. Polynomials har jagah continuous hote hain;

a

par hole ya jump wala function ise tod deta hai.

Limit laws One-sided limits par bhi utni hi acchi tarah apply hoti hain.

True. Har proof mein sirf "

x

a

ke paas" use kiya gaya;

x < a

x > a

tak restrict karne se triangle-inequality argument mein kuch nahi badalta, isliye saare chaar laws one-sidedly hold karte hain.

lim (c f) = c L

tab bhi hold karta hai jab

c = 0

ho.

True. Dono sides

0

ke barabar hain:

0 \cdot f = 0

aur

c L = 0 \cdot L = 0

, chahe

L

kuch bhi ho (chahe

L

ek finite value ke roop mein exist kare).

Error dhundo

lim_{x \to 3} \frac{x - 3}{x - 3} = \frac{0}{0}

, jo undefined hai, isliye limit exist nahi karta."

Galat conclusion. Limit sirf

x

ko 3 ke paas dekhti hai jahan

x \neq = 3

, isliye ratio poore time

1

hai aur limit

1

hai;

x = 3

par value limit ke liye irrelevant hai.

lim (f \cdot g) = lim f \cdot lim g

, isliye

lim (x \cdot \frac{1}{x}) = lim x \cdot lim \frac{1}{x} = 0 \cdot \infty

Illegal use.

lim_{x \to 0} \frac{1}{x}

(finite) exist nahi karta, isliye product law ki hypothesis fail hoti hai — tum ise apply nahi kar sakte. Pehle simplify karo:

x \cdot \frac{1}{x} = 1 \to 1

"By the quotient law

lim_{x \to 0} \frac{s i n x}{x} = \frac{l i m s i n x}{l i m x} = \frac{0}{0} = 1

Do errors hain:

0/0

koi number nahi hai, aur quotient law void hai jab

M = 0

ho. Is limit ki value 1 hai lekin iske liye Squeeze theorem ya Derivative as a limit chahiye, quotient law nahi.

lim (f + g) = L + M

, isliye main limits tab bhi add kar sakta hoon jab sirf ek exist kare."

Law dono

L

aur

M

ko existing finite limits ke roop mein name karta hai. Ek missing ho toh koi

M

nahi hai add karne ke liye, isliye equation as stated meaningless hai.

lim_{x \to a} (c f) = c L

works, isliye

lim_{x \to a} (g \cdot f) = g \cdot L

kisi bhi

g

ke liye."

Constant-multiple law ko ek genuine constant

c

chahiye, varying

g (x)

nahi. Constant ki jagah function rakhne ke liye full product law aur

lim g

ka exist karna zaroori hai.

"Sum proof mein humne

∣ f - L ∣ < ε

aur

∣ g - M ∣ < ε

choose kiya, total

< ε

deta hai."

ε

add karne se

2 ε

milta hai,

ε

nahi. Isliye parent

ε /2 + ε /2

mein split karta hai — taaki halves exactly

ε

mein reassemble hon.

Why wale questions

Product-law proof mein

L g (x)

add aur subtract kyun kiya jaata hai?

Ye total wobble ko

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

mein split karta hai — ek piece

f

ki error isolate karta hai, doosra

g

ki error — aur har piece ek known controllable limit hai jo

0

ki taraf ja rahi hai.

Product proof close hone ke liye

g

a

ke paas bounded hona kyun zaroori hai?

g (f - L)

mein, agar

g

blow up karta toh ek tiny

∣ f - L ∣

ko amplify karke kuch bada bana sakta tha. Kyunki

g \to M

, ye

a

ke paas ek fixed band mein rehta hai, isliye

g (f - L) \to 0

Quotient law par fine print sirf

M \neq = 0

kyun hai aur kahin aur nahi?

0

ki taraf jaane wali quantity se divide karna

f / g

\pm \infty

tak bhej sakta hai ya ise oscillate kara sakta hai; sirf zero-se-door denominators finite rehte hain,

1/ g \to 1/ M

prove hone deta hai.

Polynomial limits plain substitution se kyun calculate kar sakte hain?

Ek polynomial

x

ki powers ke constant-multiples ka sum hai; har atom

lim x = a

aur

lim c = c

obey karta hai, aur sum/product/const-multiple laws unhe

p (a)

mein reassemble karte hain.

Hum limit dhundhne ke liye "almost never" epsilon–delta directly kyun use karte hain?

Laws humein ek messy function ko known limits wale atoms mein decompose karne aur answers ko glue karne dete hain, isliye har law ka ek general proof countless case-by-case

ε

δ

hunts replace karta hai.

Triangle inequality sum proof mein kyun appear karta hai, aur product proof ki first line mein nahi?

Sum ki error literally

(f - L) + (g - M)

hai, do terms ka sum — exact shape

∣ p + q ∣ \leq ∣ p ∣ + ∣ q ∣

bound karta hai. Product ki error ek sum tabhi banta hai jab add-subtract trick use karne ke baad create hota hai.

Edge cases

Agar dono

L = 0

aur

M = 0

hain, toh quotient law kya batata hai?

Kuch nahi — ye apply nahi hota (

M = 0

). Tum Indeterminate forms 0 over 0 face kar rahe ho; asli limit koi bhi number,

\pm \infty

, ya nonexistent ho sakta hai.

Agar

L \neq = 0

lekin

M = 0

ho, toh kya

lim (f / g)

finite ho sakta hai?

Generally nahi — numerator ek nonzero value ki taraf ja raha hai jabki denominator vanish ho raha hai, isliye

∣ f / g ∣ \to \infty

(ya two-sided limit sign flip se fail ho jaati hai). Ye ek indeterminate form nahi hai.

Ek aisi point par jahan

lim_{x \to a^{-}} f \neq = lim_{x \to a^{+}} f

ho, kya sum/product laws kaam karti hain?

Haan, lekin sirf one-sided limits ke liye jo exist karti hain. Two-sided

lim_{x \to a} f

exist nahi karta, isliye tum sirf matching one-sided limits ko One-sided limits se combine kar sakte ho.

Kya do functions ka product jinka koi limit nahi ho, limit rakh sakta hai?

Haan.

f = sgn (x)

aur

g = sgn (x)

0

ke paas dono jump karte hain, lekin

f g \equiv 1

(

x \neq = 0

ke liye) ka limit

1

hai. Factors ka non-existence product limit ko forbid nahi karta.

Kya

lim (c f) = c L

tab bhi sense banta hai agar

L

+ \infty

ho?

Stated law assume karta hai

L

finite ho.

L = + \infty

ke saath tum finite-limit laws se bahaar ja rahe ho; extended-real rules apply hote hain (aur

c = 0

tab indeterminate

0 \cdot \infty

deta hai).

Sum law ka kya hoga agar ek limit

+ \infty

ho aur doosri

- \infty

ho?

Finite law ise cover nahi karta;

\infty - \infty

indeterminate hai. Tumhe specific functions ko directly analyse karna hoga — algebraic law koi jawab nahi deta.

Connections

Limit laws — sum, product, quotient, constant multiple — parent jiska fine print ye traps probe karte hain.
Epsilon-Delta definition of a limit — upar har "kyun" wale jawab ka source.
One-sided limits — jahan mismatched left/right limits two-sided combination tod dete hain.
Indeterminate forms 0 over 0 — destination jab quotient law void ho jaata hai.
Continuity — isliye direct substitution sirf continuous functions ke liye kaam karta hai.
Squeeze theorem — un product/quotient limits ko bachata hai jo algebra laws nahi kar sakte.
Derivative as a limit — jahan ye same traps difference quotient mein phir appear hote hain.