Limits and continuity in 2D — path-dependence issue: Question bank

4.4.2 · D5 · Maths › Multivariable Calculus › 2D mein Limits aur Continuity — path-dependence issue

Shuru karne se pehle, vocabulary ka ek reminder taaki yahan kuch bhi unexplained na rahe:

Ek path koi bhi curve hai (line, parabola, spiral) jo target point $(a, b)$ par khatam hoti hai.
Limit exist karti hai tabhi jab har path same value $L$ de — kyunki $ε$ – $δ$ definition demand karti hai ki poora disk kaam kare, sirf kuch roads nahi.
Matching paths sirf limit ko DISprove kar sakti hain (ek mismatch dhundh kar). Existence prove karne ke liye Squeeze Theorem ya $ε$ – $δ$ bound chahiye.

True ya false — justify karo

Agar origin se do straight-line paths same value dein, toh limit exist karti hai.

False. Do — ya infinitely many lines bhi — coincidence se agree kar sakti hain;

\frac{x y ^{2}}{x ^{2} + y ^{4}}

har line par

0

deta hai lekin

x = y^{2}

ke along

\frac{y ^{2} \cdot y ^{2}}{( y ^{2} ) ^{2} + y ^{4}} = \frac{y ^{4}}{y ^{4} + y ^{4}} = \frac{y ^{4}}{2 y ^{4}} = \frac{1}{2}

ho jaata hai. Agreement sirf limits ko rule out karti hai, kabhi rule in nahi karti.

Agar har straight line ke along limit exist karti hai, toh limit exist karti hai.

False. "Har line" phir bhi curved paths jaise parabolas aur spirals ko miss karti hai; balanced curve

x = y^{2}

classic escapee hai. Lines jhooth bolti hain, curves sach bolta hain.

Agar ek path

0

deti hai aur doosri

\frac{1}{2}

, toh limit exist nahi karti.

True. Do paths ke beech ek genuine disagreement "all paths agree" ko violate karti hai, isliye koi common

L

exist nahi kar sakta. Yahi killer test ki poori power hai.

Degree

0

ka homogeneous function origin par limit rakh sakta hai.

Kabhi-kabhi, haan: sirf tabhi jab yeh saari rays par constant ho (har direction mein same value). Agar koi bhi do rays differ karein, toh iska directional constancy guarantee karta hai ki limit nahi hai. Aam taur par yeh fail hota hai.

Agar

f

ki polar form mein

r \to 0

par

θ

bacha rahe, toh limit exist nahi karti.

Precisely true tabhi jab

θ

-terms nahi kill hote kisi

\to 0

factor se. Clean criterion yeh hai:

f = g (r, θ)

likho; agar tum

∣ f - L ∣ \leq h (r)

bound kar sako jahan

h (r) \to 0

θ

se independent ho, toh limit

L

hai. Agar balki

θ

-dependence

r \to 0

par survive kare (jaise

cos θ sin θ

, sirf

θ

ka function bina kisi shrinking

r

-factor ke), toh limit exist nahi karti. Toh

r cos θ

harmless hai (

r \to 0

ise crush kar deta hai); akela

cos θ sin θ

fatal hai.

(a, b)

par continuity ke liye sirf limit ka exist karna zaroori hai.

False. Iske liye teen cheezein chahiye:

f (a, b)

exist kare, limit exist kare, aur woh equal hon. Point par ek hole ya jump continuity tod deta hai chahe limit exist bhi kare.

Agar

lim_{(x, y) \to (0, 0)} f = L

exist karta hai, toh

y = x^{2}

ke along approach bhi

L

degi.

True. Existence ka matlab hai ki har path

L

deti hai, isliye koi bhi specific path — parabola bhi — zaroor agree karegi. Isliye ek deviating path fatal hoti hai.

∣ f (x, y) ∣ \leq ∣ y ∣

origin ke paas hona

lim f = 0

conclude karne ke liye kaafi hai.

True, bas yeh bound poore disk par hold kare: jaise

∣ y ∣ \to 0

, Squeeze Theorem

f \to 0

ko har path par force karta hai ek saath, jo exactly wahi hai jo

ε

–

δ

ko chahiye.

Error dhundho

"Maine

y = 0

aur

x = 0

check kiya; dono ne

0

diya, toh limit

0

hai."

Error yeh hai ki do paths se existence claim ki ja rahi hai. Do axes sirf do directions hain infinitely many mein se; woh sirf ek guess support kar sakte hain, kabhi proof nahi — slopes aur ek parabola try karo, phir Squeeze.

y = m x

ke along (jahan

m \in R

slope hai),

\frac{x y}{x ^{2} + y ^{2}} = \frac{m}{1 + m ^{2}}

;

m = 0

plug karne se

0

aata hai, toh limit

0

hai."

Value

\frac{m}{1 + m ^{2}}

$m$ par depend karti hai — woh dependence khud hi proof hai ki limit fail hoti hai (

m = 1

\frac{1}{2}

deta hai). Ek single

m

read karna disagreement ko hide kar deta hai.

"Polar mein,

\frac{x y}{x ^{2} + y ^{2}} = cos θ sin θ \to 0

jaise

r \to 0

Koi

r

nahi hai jo

0

par jaaye; expression pure $θ$ hai. Iska value

cos θ sin θ

hamesha rehta hai, toh yeh kabhi settle nahi hota — limit exist nahi karti.

f (y^{2}, y) = \frac{y ^{4}}{2 y ^{4}} = \frac{1}{2}

, aur

\frac{1}{2} \neq = 0

, toh

f

discontinuous hai — lekin sirf us ek curve par."

Discontinuity ek property hai point par, "curve par" nahi. Mismatch ka matlab hai ki origin par koi bhi value

L

exist nahi karti, isliye

f

ko koi bhi assignment dekar wahan continuous nahi banaya ja sakta.

"Limit exist karti hai kyunki main

∣ f ∣ \leq r^{2} cos θ

bound kar sakta hoon."

Valid bound direction par depend nahi karni chahiye;

r^{2} cos θ

abhi bhi

θ

carry karta hai. Balki akele

r^{2}

se bound karo (kyunki

∣ cos θ ∣ \leq 1

) taaki yeh uniformly

0

par shrink kare.

"Kyunki

(0, 0)

par har partial derivative exist karti hai,

f

wahan continuous hai."

Partial Derivatives sirf do axis directions probe karti hain; unka exist karna curved approaches ke baare mein kuch nahi kehta. Ek function dono partials rakh sakta hai phir bhi continuity fail kar sakta hai (path-dependent limit).

Why questions

ε

–

δ

condition ek disk kyun use karti hai, line kyun nahi?

Disk output ko

L

ke paas saari directions se simultaneously force karti hai, encode karke "har path agree karti hai." Ek line sirf ek approach control karegi aur path-dependence bilkul miss kar degi.

Matching paths kabhi bhi limit exist karna prove kyun nahi kar sakti?

Tum infinitely many paths mein se sirf countable handful hi test kar sakte ho; untested wali (jaise balanced curve) disagree kar sakti hain. Mismatch nahi mila, yeh proof nahi hai ki koi nahi hai.

Polar Coordinates mein convert karna origin par kyun help karta hai?

Yeh "

(x, y) \to (0, 0)

" ko single condition "

r \to 0

" mein badal deta hai, kitna door (

r

) aur kaunsi direction (

θ

) ko alag kar deta hai. Agar

θ

phir bound se vanish ho jaaye, saari directions ek value par collapse ho jaati hain.

\frac{x y}{x ^{2} + y ^{2}}

origin se har ray ke along constant kyun hai?

Yeh degree

0

ka homogeneous hai:

(x, y) \to (t x, t y)

scale karne se yeh unchanged rehta hai, isliye iska value sirf ray ki direction par depend karta hai, distance par nahi. Isliye alag rays alag constants lock kar leti hain.

Jab saari lines agree karein toh specifically parabola

x = y^{2}

kyun try karte hain?

Yeh numerator aur denominator ko same order banata hai (

\frac{x y ^{2}}{x ^{2} + y ^{4}}

mein dono

y^{4}

ban jaate hain), isliye offending term ab vanish nahi hota — exactly wahi "balanced" path jo lines nahi reach kar sakti.

2D limit exist karna prove karne ke liye Squeeze Theorem sahi tool kyun hai?

Yeh "har path check karo" ko ek single inequality

∣ f - L ∣ \leq g (r) \to 0

se replace karta hai jo poore disk par hold kare, jo saari paths ko ek saath certify karta hai — directly

ε

–

δ

requirement match karta hai.

Naive line test

y = m x

automatically parabola trap kyun nahi pakad paata?

Lines

y

x

ke proportional force karti hain, isliye

y^{4} = m^{4} x^{4}

x^{2}

se higher order rehta hai aur mar jaata hai; sirf tabhi jab

x

y^{2}

ki tarah scale kare, orders match hote hain aur hidden value appear hoti hai.

Edge cases

Agar

f

sirf

(0, 0)

par undefined hai lekin limit exist karti hai, toh kya hoga?

Function mein ek removable discontinuity hai:

f (0, 0) = L

define karo aur yeh continuous ho jaata hai. Limit ka exist karna exactly wahi cheez hai jo repair possible banati hai.

(0, 0)

ke approach mein constant path "origin par hi raho" — kya yeh allowed hai?

Nahi. Definition require karti hai

0 < x^{2} + y^{2}

, yaani strictly point se door; point par ki value limit mein kabhi enter nahi karti, sirf surrounding disk karta hai.

Agar

f \equiv 3

har jagah hai sivaay ek single ray ke jahan

f = 5

hai, toh kya origin par limit exist karti hai?

Nahi. Us ray ke along approach

5

deti hai jabki har doosri direction

3

deti hai; ek deviating direction, chahe kitni bhi "thin" ho, agreement tod deti hai aur limit kill kar deti hai.

Kya har parabola

y = c x^{2}

ke along limit exist kar sakti hai lekin phir bhi fail ho sakti hai?

Haan. Jaise lines parabolas miss karti hain, parabolas doosri curves miss kar sakti hain (jaise

y = x^{3}

ya spirals). Paths ka koi bhi finite family kabhi complete nahi hota — sirf ek uniform bound settle karta hai.

Kya hoga agar polar bound

∣ f ∣ \leq r

ho lekin sirf

θ \neq = \frac{π}{2}

ke liye?

Bound uniform nahi hai: yeh ek direction par fail karti hai, toh tumne poore disk ko control nahi kiya. Tumhe excluded direction alag se handle karna hoga ya "proof" mein hole hai.

Ek spiral

r = θ

ke along origin mein,

\frac{x y}{x ^{2} + y ^{2}}

ki limit hai?

Nahi. Spiral par

θ

bina settle kiye continuously change karta rehta hai, isliye

cos θ sin θ

andar tak oscillate karta hai — spiral kaafi directions sample karti hai aur kabhi converge nahi karti.

Agar ek restriction ke dono one-sided 1D limits match karein, toh kya 2D limit safe hai?

Nahi. Ek restriction ek single path hai; uske do "sides" 2D mein approach ki sirf ek direction hain. Yeh 1D confidence borrow karta hai jo simply plane mein transfer nahi hoti.

Connections

Multivariable Calculus — parent chapter
Squeeze Theorem — existence prove karne ka ek hi reliable tarika
Polar Coordinates — distance $r$ ko direction $θ$ se alag karta hai
Partial Derivatives — sirf axis limits, continuity ke liye insufficient
Differentiability in 2D — continuity demand karta hai, isliye path-independence bhi
Epsilon-Delta Definition — disk condition jo sab kuch underlie karti hai
Continuity in 1D — jahan se misleading two-sided intuition aati hai