Worked examples — Limits and continuity in 2D — path-dependence issue

4.4.2 · D3 · Maths › Multivariable Calculus › Limits and continuity in 2D — path-dependence issue

Yeh page ek drill floor hai. Parent note ne idea build kiya; yahan hum har tarah ke case ko cover karte hain jo yeh topic throw kar sakta hai, ek worked example per cell. Har solution se pehle ek Forecast line hai — answer cover karo aur pehle guess karo. Yahi aadat poora game hai.

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Scenario matrix

Har 2D-limit question jo tum kabhi bhi miloge, exactly inhi boxes mein se ek mein aata hai. Hamare neeche ke examples inhe sab cover karte hain.

Cell	Ise woh cell kya banata hai	Killer tool	Example
A. Line test khatam karta hai	Do straight lines pehle se disagree karte hain	Do paths	Ex 1
B. Lines jhooth bolti hain, curve confess karta hai	Saari lines agree karti hain, ek parabola/cubic tod deta hai	Smart curve	Ex 2
C. Genuinely exist karta hai (numerator jeet jaata hai)	Top degree bottom ko beat karti hai → shr ink hota hai	Squeeze / polar	Ex 3
D. Sign / quadrant sensitivity	Value flip ho jaati hai depending on kis quadrant se enter kar rahe ho	Signed paths	Ex 4
E. Degenerate / removable-looking	$0/0$ lagta hai lekin actually harmless hai	Algebra + polar	Ex 5
F. Non-origin target point	Ek aisi point $(a,b)\ne (0,0)$ tak approach karo	Shift coordinates	Ex 6
G. Real-world word problem	Physical quantity, DNE interpret karna zaroori hai	Model + paths	Ex 7
H. Exam twist: $k$ choose karo taaki continuous ho	Woh value solve karo jo hole ko patch karti hai	Limit = value	Ex 8

Task ke columns "signs/quadrants", "zero/degenerate inputs", "limiting values", "word problem" aur "exam twist" Cells D, E, C, G, H hain respectively. Chalte hain inhe dekhte hain.

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Cell A — line test akela hi khatam karta hai

Forecast: Top aur bottom dono degree 2 hain — same "size." Jab top aur bottom saath scale karte hain, to $x$'s aur $y$'s usually cancel nahi hote, isliye mujhe lagta hai answer direction par depend karta hai → DNE.

Step 1 — $x$-axis ke along approach karo ($y=0$). $f(x,0)=\frac{x^2-0}{x^2+0}=1\to 1.$ Yeh step kyun? $y=0$ set karna sabse sasta path hai — yeh do-variable ka mess ek variable mein collapse kar deta hai. $x$-axis $1$ deta hai.

Step 2 — $y$-axis ke along approach karo ($x=0$). $f(0,y)=\frac{0-y^2}{0+y^2}=-1\to -1.$ Yeh step kyun? Doosra free axis next sabse sasta path hai. Yeh pehle hi $-1$ deta hai.

Step 3 — Compare karo. $x$-axis wala path $1$ kehta hai; $y$-axis wala path $-1$ kehta hai. Do paths, do answers.

Verify (polar sanity check): $x=r\cos \theta ,y=r\sin \theta$ ke saath, $\frac{r^2{\cos }^2\theta -r^2{\sin }^2\theta }{r^2}={\cos }^2\theta -{\sin }^2\theta =\cos 2\theta .$ $r$ completely cancel ho gaya — pure direction $\theta$ bacha. $\theta =0$ gives $1$, $\theta =\frac{\pi }{2}$ gives $-1$. Confirmed: limit DNE.

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Cell B — lines jhooth bolti hain, curve confess karta hai

Forecast: Yeh parent ke $\frac{x{y}^2}{x^2+y^4}$ trap ka cousin hai. Mujhe predict karta hoon ki har straight line $0$ degi, lekin ek sahi choose ki gayi parabola ise tod degi.

Step 1 — Har straight line $y=mx$. $f(x,mx)=\frac{x^2\cdot mx}{x^4+m^2{x}^2}=\frac{m{x}^3}{x^2(x^2+m^2)}=\frac{mx}{x^2+m^2}\underset{x\to 0}{\to }\frac{0}{m^2}=0.$ Yeh step kyun? Denominator se $x^2$ factor karo; top par ek akela $x$ bachta hai aur poori cheez $0$ par le jaata hai. Har line $0$ kehti hai ($m\ne 0$ ke liye; axes bhi inspection se $0$ deti hain).

Step 2 — Balanced curve $y=x^2$ try karo. Hum chahte hain ki numerator aur denominator same order ke hon. Numerator $x^{2}y$ ban jaata hai $x^2\cdot x^2=x^4$; denominator ban jaata hai $x^4+x^4$. Match! $f(x,x^2)=\frac{x^2\cdot x^2}{x^4+(x^2{)}^2}=\frac{x^4}{x^4+x^4}=\frac{x^4}{2x^4}=\frac{1}{2}.$ Yeh step kyun? Lines us direction ko miss kar gayi jahan top aur bottom same rate se shrink karte hain. Yahi woh curve hai jise hamesha dhundhna chahiye.

Verify: lines $\to 0$, parabola $\to \frac{1}{2}$. Kyunki $0\ne \frac{1}{2}$, limit DNE. ("Lines Lie, Curves Confess" mnemonic in action.)

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Cell C — yeh genuinely exist karta hai (limiting value)

Forecast: Numerator degree 4 hai, denominator degree 2. Jaise $r^4/r^2=r^2\to 0$. Mujhe lagta hai limit $0$ hai — aur paths ise prove nahi kar sakte, isliye mujhe Squeeze ya polar chahiye hoga.

Step 1 — Polar mein convert karo. $x=r\cos \theta ,y=r\sin \theta$ set karo. $\frac{r^4{\cos }^4\theta +r^4{\sin }^4\theta }{r^2}=r^2({\cos }^4\theta +{\sin }^4\theta ).$ Yeh step kyun? Polar "kitna close hai" ($r$) ko "kis direction se" ($\theta$) se alag kar deta hai. Agar $r\to 0$ par bhi answer mein $\theta$ hai, to yeh path-dependent hai; agar $\theta$ sirf ek bounded factor ke andar baith jaata hai, to hum ise squeeze kar sakte hain.

Step 2 — Direction factor ko bound karo. Kyunki ${\cos }^4\theta \le 1$ aur ${\sin }^4\theta \le 1$, $0\le {\cos }^4\theta +{\sin }^4\theta \le 2.$ Yeh step kyun? Hume trig part ki exact value nahi chahiye — sirf yeh ki yeh blow up nahi kar sakta. Ise bound karna saari direction-dependence khatam kar deta hai.

Step 3 — Squeeze. $0\le \left|\frac{x^4+y^4}{x^2+y^2}\right|=r^2({\cos }^4\theta +{\sin }^4\theta )\le 2r^2.$ Jab $(x,y)\to (0,0)$, $r\to 0$, to $2r^2\to 0$. Dono walls $0$ par collapse ho jaate hain. Squeeze Theorem se middle poore disk par $0$ hone par force ho jaata hai — koi bhi path escape nahi kar sakta. $\boxed{\underset{(x,y)\to (0,0)}{\lim }\frac{x^4+y^4}{x^2+y^2}=0.}$

Verify: koi bhi line $y=mx$ try karo: $\frac{x^4+m^4{x}^4}{x^2+m^2{x}^2}=\frac{x^2(1+m^4)}{1+m^2}\to 0$. Boxed answer se consistent.

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Cell D — sign / quadrant sensitivity

Forecast: $|x|$ (absolute value of $x$) ka matlab hai $x$ ka sign wipe out ho jaata hai lekin $y$ ka sign bachta hai. $y=mx$ ke along mujhe expect hai ki value abhi bhi slope ke sign par depend karegi → DNE, aur $|x|$ ki wajah se quadrants I & II ek jaisi behave karengi lekin III & IV ke against flip ho jayegi.

Step 1 — Line $y=mx$ with $x>0$ (to $|x|=x$). $f(x,mx)=\frac{x\cdot mx}{x^2+m^2{x}^2}=\frac{m{x}^2}{x^2(1+m^2)}=\frac{m}{1+m^2}.$ Yeh step kyun? $x>0$ fix karo taaki $|x|$ sirf $x$ ho; phir yeh parent ke Example-1 algebra jaisa hai. Slope $m$ bachta hai → pehle se directional.

Step 2 — Do directions test karo. $m=0$ gives $0$; $m=1$ gives $\frac{1}{2}$. Different → DNE already.

Step 3 — Quadrant story dekho (kyun $|x|$ matter karta hai). Left se enter karo, $x<0$, $y=mx$ ke along to ab $|x|=-x$: $f(x,mx)=\frac{(-x)(mx)}{x^2(1+m^2)}=\frac{-m{x}^2}{x^2(1+m^2)}=\frac{-m}{1+m^2}.$ Yeh step kyun? $x<0$ ke saath, $|x|$ numerator ka sign flip kar deta hai. To slope $m$ ke saath right se approach karne par $\frac{m}{1+m^2}$ milta hai lekin left se $\frac{-m}{1+m^2}$ milta hai — usi geometric line ki do half-lines disagree karti hain! Yeh pure quadrant sensitivity hai.

Verify (polar): $|x|=r|\cos \theta |$, to $\frac{r|\cos \theta |\cdot r\sin \theta }{r^2}=|\cos \theta |\sin \theta ,$ pure $\theta$. $\theta =\frac{\pi }{4}$ par: $\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=\frac{1}{2}$. $\theta =\frac{3\pi }{4}$ par (quadrant II): $|-\frac{1}{\sqrt{2}}|\cdot \frac{1}{\sqrt{2}}=\frac{1}{2}$ — I jaisa hi. $\theta =-\frac{\pi }{4}$ par (quadrant IV): $\frac{1}{\sqrt{2}}\cdot (-\frac{1}{\sqrt{2}})=-\frac{1}{2}$. Upper half $+\frac{1}{2}$, lower half $-\frac{1}{2}$ → DNE. Confirmed.

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Cell E — degenerate-looking lekin harmless

Forecast: $(0,0)$ par yeh $\frac{0}{0}$ hai — woh classic "degenerate" indeterminate form jo logon ko daraa deta hai. Lekin numerator degree 4, denominator degree 2 ko beat karta hai, isliye mujhe expect hai yeh vanish ho jaayega: limit $=0$.

Step 1 — Ek factor ko bound karo. Notice karo ki $\frac{y^2}{x^2+y^2}\le 1$ kyunki denominator kam se kam $y^2$ hai. Yeh step kyun? Hum ek aisa piece peel off karna chahte hain jo provably shrink ho aur baaki ko $1$ se cap kar den.

Step 2 — Rewrite aur bound karo. $0\le \frac{x^2{y}^2}{x^2+y^2}=x^2\cdot \frac{y^2}{x^2+y^2}\le x^2.$ Yeh step kyun? $x^2$ pull out karo (jo clearly $\to 0$ hai) aur remaining fraction ko $1$ se cap karo. Woh messy denominator ab tame ho gaya.

Step 3 — Squeeze. Jab $(x,y)\to (0,0)$, $x^2\to 0$, to middle $0$ par trap ho jaata hai. $\boxed{\underset{(x,y)\to (0,0)}{\lim }\frac{x^2{y}^2}{x^2+y^2}=0.}$

Verify (polar): $\frac{r^2{\cos }^2\theta r^2{\sin }^2\theta }{r^2}=r^2{\cos }^2\theta {\sin }^2\theta$, aur $|{\cos }^2\theta {\sin }^2\theta |\le \frac{1}{4}$, isliye $|f|\le \frac{1}{4}{r}^2\to 0$. "$0/0$" kabhi real problem nahi tha — top bas jeet jaata hai.

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Cell F — target point jo origin nahi hai

Forecast: Yeh Example-1 ka DNE structure ($\frac{uv}{u^2+v^2}$) hai lekin origin ki jagah $(1,2)$ par centred hai. Mujhe bet hai ek coordinate shift ise exact same "slope par depend karta hai" wali failure mein turn kar deta hai → DNE.

Step 1 — Coordinates shift karo. $u=x-1,v=y-2$ lo. Yeh step kyun? $\epsilon$–$\delta$ definition sirf target se distance ki parwah karta hai. Poore plane ko slide karna taaki $(1,2)$ origin par aa jaaye, limit ke baare mein kuch nahi badalta lekin algebra familiar ho jaata hai. Jab $(x,y)\to (1,2)$, hume $(u,v)\to (0,0)$ milta hai.

Step 2 — Rewrite karo. $\frac{(x-1)(y-2)}{(x-1{)}^2+(y-2{)}^2}=\frac{uv}{u^2+v^2}.$ Yeh step kyun? Parent Example 1 se identical. Hum already prove kar chuke hain ki yeh directional hai.

Step 3 — Known result reuse karo. $v=mu$ ke along: value $=\frac{m}{1+m^2}$, jo $m$ ke saath change hoti hai ($m=0\to 0$, $m=1\to \frac{1}{2}$).

Verify: limit $(1,2)$ par DNE — hole shifted origin par hai, exactly expected jaisa. (Sanity: path $y=2$, yaani $v=0$ plug karo: $\frac{0}{u^2}=0$; $y-2=x-1$, yaani $v=u$ plug karo: $\frac{u^2}{2u^2}=\frac{1}{2}$. Mismatch confirmed.)

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Cell G — real-world word problem

Forecast: Yeh Example 1 hai jo physics ka costume pehne hua hai. Agar mathematical limit DNE hai, to physically koi bhi consistent temperature hole ko fill nahi kar sakti — plate ke centre mein reading ka ek genuine "crack" hai.

Step 1 — Limit ke roop mein restate karo. "Origin par smooth reading" ka matlab hai $\underset{(x,y)\to (0,0)}{\lim }T(x,y)$ exist karna chahiye aur hum $T(0,0)$ ko uske equal set karte hain (parent note se continuity conditions). Yeh step kyun? Continuity = limit exist kare aur assigned value ke equal ho. Pehle hume jaanna hoga ki limit exist bhi karti hai.

Step 2 — Centre tak do physical routes se chalo. Origin tak wire $y=0$ ke along approach karo (plate ke aarpaar ek seedha scratch): $T(x,0)=\frac{0}{x^2}=0^{\circ }$C. Diagonal $y=x$ ke along approach karo: $T(x,x)=\frac{x^2}{2x^2}=0.5^{\circ }$C. Yeh step kyun? Do physical paths, usi point par approach karte hue do alag temperatures.

Step 3 — Interpret karo. Kyunki scratch ke along slide karne wala sensor $0^{\circ }$ read karega lekin diagonally slide karne wala $0.5^{\circ }$ read karega, koi bhi single value hole ko patch nahi kar sakti. Limit DNE hai, isliye plate centre par unavoidably discontinuous hai.

Verify (units + polar): $T=\cos \theta \sin \theta =\frac{1}{2}\sin 2\theta$ °C — ek dimensionless ratio times °C, units sahi hain. Iski value sirf direction se $[-\frac{1}{2},\frac{1}{2}]$ °C tak range karti hai; e.g. $\theta =\frac{\pi }{4}\Rightarrow 0.5$°C, $\theta =0\Rightarrow 0$°C. Koi consistent origin temperature exist nahi karti.

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Cell H — exam twist: ise continuous banane ke liye $k$ choose karo

Forecast: Parent ke Forecast-then-Verify drill ne pehle hi is fraction ko handle kiya tha: degree 3 over degree 2 $r\to 0$ jaisi behave karti hai, isliye limit $0$ hai. Agar limit exist karti hai, to patching value uske equal honi chahiye — isliye mujhe predict karta hoon $k=0$.

Step 1 — Limit compute karo (polar). $\frac{r^3({\cos }^3\theta +{\sin }^3\theta )}{r^2}=r({\cos }^3\theta +{\sin }^3\theta ).$ Yeh step kyun? Continuity limit maangti hai; polar ise cleanly deta hai aur koi bhi direction-dependence expose karta hai.

Step 2 — Bound karo aur squeeze karo. Kyunki $|{\cos }^3\theta +{\sin }^3\theta |\le 2$, $0\le |f|\le 2r\underset{r\to 0}{\to }0.$ To $\underset{(x,y)\to (0,0)}{\lim }f=0$ — direction-independent, limit exist karti hai. Yeh step kyun? Bounded trig factor ka matlab hai koi bhi path disagree nahi karta; limit genuinely $0$ hai.

Step 3 — Value ko limit se match karo. Continuity ke liye $f(0,0)=\lim f$ chahiye. Isliye $\boxed{k=0.}$ Koi bhi doosra $k$ ek jump chhodta hai: surrounding values $0$ par approach karti hain lekin assigned centre $k\ne 0$ par baithega — ek discontinuity.

Verify: $k=0$ ke saath, continuity ki teeno conditions hold karti hain — $f(0,0)=0$ exist karta hai, limit $=0$ exist karti hai, aur dono equal hain. Line check $y=x$: $\frac{2x^3}{2x^2}=x\to 0$. ✔

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Case-coverage recap

Recall Kya humne har cell hit kiya?

Cell A (line test) ::: Ex 1 — $\frac{x^2-y^2}{x^2+y^2}$ DNE. Cell B (curve confesses) ::: Ex 2 — $\frac{x^{2}y}{x^4+y^2}$ DNE via $y=x^2$. Cell C (exists, limiting value) ::: Ex 3 — $\frac{x^4+y^4}{x^2+y^2}=0$ by Squeeze. Cell D (sign/quadrant) ::: Ex 4 — $\frac{|x|y}{x^2+y^2}$ upper/lower half ke beech flip karta hai. Cell E (degenerate 0/0) ::: Ex 5 — $\frac{x^2{y}^2}{x^2+y^2}=0$, harmless. Cell F (non-origin point) ::: Ex 6 — origin par shift karo, $(1,2)$ par DNE. Cell G (word problem) ::: Ex 7 — heated plate, koi patchable centre temperature nahi. Cell H (choose $k$) ::: Ex 8 — $k=0$ ise continuous banata hai.

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Connections

• Multivariable Calculus — parent chapter
• Parent topic note
• Squeeze Theorem — Cells C, E, H ko prove karta hai
• Polar Coordinates — har "exists?" verdict ke peeche $r,\theta$ split
• Epsilon-Delta Definition — kyun poora disk matter karta hai, sirf paths nahi
• Continuity in 1D — woh two-sided intuition jo hum yahan generalise kar rahe hain
• Partial Derivatives — axis-direction limits wahan dobara aate hain
• Differentiability in 2D — continuity ki zaroorat hai jo hum is page par test karte hain