Limits and continuity in 2D — path-dependence issue

4.4.2 · Maths › Multivariable Calculus

HUM KYA POOCH RAHE HAIN

PATH-DEPENDENCE LIMIT KO KAISE KHATAM KARTA HAI

Try karne wale standard paths, order mein (sabse saste pehle):

$x$ -axis ke along: $y = b$ set karo, $x \to a$ karo.
$y$ -axis ke along: $x = a$ set karo, $y \to b$ karo.
Lines $y = b + m (x - a)$ ke along general slope $m$ ke liye.
Parabolas jaise $y = (x - a)^{2}$ ke along agar lines sab agree karein lekin aapko shak ho.

Worked Example 1 — canonical failure

$lim_{(x, y) \to (0, 0)} \frac{x y}{x ^{2} + y ^{2}}$

Path A — $x$ -axis ke along ( $y = 0$ ): $f (x, 0) = \frac{x \cdot 0}{x ^{2} + 0} = 0 \to 0.$ Yeh step kyun? $y = 0$ set karne se problem 1D mein collapse ho jaati hai; numerator zero ho jaata hai.

Path B — line $y = m x$ ke along: $f (x, m x) = \frac{x \cdot m x}{x ^{2} + m ^{2} x ^{2}} = \frac{m x ^{2}}{x ^{2} ( 1 + m ^{2} )} = \frac{m}{1 + m ^{2}} .$ Yeh step kyun? $y = m x$ substitute karo aur $x^{2}$ cancel karo (valid hai kyunki path pe $x \neq = 0$ ). $x$ vanish ho jaata hai — value sirf slope $m$ par depend karti hai.

Verdict: $m = 0$ se $0$ milta hai; $m = 1$ se $\frac{1}{2}$ milta hai. Alag paths → alag values → limit does not exist.

Worked Example 2 — lines jhooth bolti hain, parabola sach batati hai

$lim_{(x, y) \to (0, 0)} \frac{x y ^{2}}{x ^{2} + y ^{4}}$

Saari straight lines $y = m x$ : $f (x, m x) = \frac{x \cdot m ^{2} x ^{2}}{x ^{2} + m ^{4} x ^{4}} = \frac{m ^{2} x ^{3}}{x ^{2} ( 1 + m ^{4} x ^{2} )} = \frac{m ^{2} x}{1 + m ^{4} x ^{2}} \to 0.$ Yeh step kyun? Denominator se $x^{2}$ factor karo; numerator mein bacha hua $x$ force karta hai limit ko $0$ ki taraf jab $x \to 0$ . Toh har line $0$ bolti hai.

Ab curve $x = y^{2}$ : $f (y^{2}, y) = \frac{y ^{2} \cdot y ^{2}}{( y ^{2} ) ^{2} + y ^{4}} = \frac{y ^{4}}{y ^{4} + y ^{4}} = \frac{y ^{4}}{2 y ^{4}} = \frac{1}{2} .$ Yeh step kyun? Humne $x = y^{2}$ choose kiya taaki numerator aur denominator same order ke ho jaayein ( $y^{4}$ ). Yeh woh "balanced" path hai jo lines miss kar gayi.

Verdict: lines $0$ deti hain, parabola $\frac{1}{2}$ deti hai → limit does not exist. (Yeh wahi steel-man trap hai jo concrete form mein hai.)

Worked Example 3 — limit DOES exist (Squeeze proof)

$lim_{(x, y) \to (0, 0)} \frac{x ^{2} y}{x ^{2} + y ^{2}}$

Aap yeh paths se prove nahi kar sakte. Ek bound use karo.

Step 1: Note karo ki $\frac{x ^{2}}{x ^{2} + y ^{2}} \leq 1$ sabhi $(x, y) \neq = (0, 0)$ ke liye. Kyun? Denominator $x^{2} + y^{2} \geq x^{2}$ , toh fraction zyada se zyada 1 hai.

Step 2: Isliye $\frac{x ^{2} y}{x ^{2} + y ^{2}} = ∣ y ∣ \cdot \frac{x ^{2}}{x ^{2} + y ^{2}} \leq ∣ y ∣.$ Yeh step kyun? $∣ y ∣$ bahar nikalo, phir baaki fraction ko 1 se bound karo.

Step 3: Jab $(x, y) \to (0, 0)$ , $∣ y ∣ \to 0$ . Squeeze Theorem se poora expression $\to 0$ . $(x, y) \to (0, 0) lim \frac{x ^{2} y}{x ^{2} + y ^{2}} = 0$ Yeh airtight kyun hai: bound $∣ f ∣ \leq ∣ y ∣$ poore disk pe hold karta hai, toh koi bhi path isse escape nahi kar sakta. Exactly yahi $ε$ – $δ$ ko chahiye.

2D mein Continuity

Forecast-then-Verify drill

Flashcards

2D limit exist karne ke liye kya hona chahiye?

Point ke paas jaane wala har path same value dena chahiye (poora disk, sirf do directions nahi).

Kya finitely many paths match karna 2D limit exist hone ko prove kar sakta hai?

Nahi — yeh sirf isse disprove kar sakta hai (mismatch dhundhke). Proof ke liye

ε

δ

ya Squeeze chahiye.

Origin pe

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

ki limit?

Does not exist;

y = m x

ke along yeh

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

ke barabar hai, jo

m

ke saath vary karta hai.

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

students ko kyun trap karta hai?

Saari straight lines

0

deti hain, lekin path

x = y^{2}

\frac{1}{2}

deta hai → limit DNE.

Polar form mein

r \to 0

ke baad bhi

θ

bachne ka kya matlab hai?

Value direction pe depend karti hai → path-dependent → limit does not exist.

(a, b)

f

ki continuity ke 3 conditions batao.

f (a, b)

exist kare; limit exist kare; limit

f (a, b)

ke barabar ho.

Origin pe

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

ki limit aur kyun?

0

; poore disk pe Squeeze se

∣ f ∣ \leq ∣ y ∣ \to 0

bound karo.

Degree 0 ka homogeneous function — iske directional limits mein kya special hai?

Har ray ke along constant, toh alag rays generally alag values deti hain → usually koi limit nahi.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum aur tumhare doston ek foggy field mein ek lamppost ki taraf chal rahe ho. 1D mein tum sirf left road ya right road se aa sakte ho. Lekin ek field mein tum kisi bhi direction se aa sakte ho — seedha, tedha, zig-zag. "Limit exist karna" ka matlab hai: chahe tum kisi bhi path se chalo, lamppost ke bilkul paas zameen ki height sabko same dikhni chahiye. Agar tumhare doston ko curvy path pe zameen height 0 dikhti hai lekin tumhe straight path pe height ½ dikhti hai, toh lamppost ke paas ek ajeeb crack hai — "limit" toot gayi. Yeh prove karne ke liye ki yeh smooth hai, tumhe dikhana hoga ki sabhi log, har imaginable path pe, same height pe pahunchte hain — usually height ko do squeezing walls ke beech trap karke jo dono same number tak shrink hoti hain.

Connections

Multivariable Calculus — parent chapter
Squeeze Theorem — 2D limit exist hone ko prove karne ka tool
Polar Coordinates — $(x, y) \to (0, 0)$ ko $r \to 0$ mein convert karta hai clean test ke liye
Partial Derivatives — axis-directions ke along limits pe built hai
Differentiability in 2D — continuity chahiye, jo path-independent limits maangti hai
Epsilon-Delta Definition — 1D ancestor jo disks tak generalize hua
Continuity in 1D — contrast: sirf do directions vs. infinitely many