Exercises — Limits and continuity in 2D — path-dependence issue

$x$-axis ke along $y=0$ rakh do: $f(x,0)=\frac{x-0}{x+0}=1\to 1.$ $y$-axis ke along $x=0$ rakh do: $f(0,y)=\frac{0-y}{0+y}=-1\to -1.$ Humne kya kiya: do baar 1D mein collapse kiya. Kyun: sabse saste paths pehle. Yeh $1$ aur $-1$ dete hain. Verdict: do paths disagree karte hain $\Rightarrow$ limit DNE.

Sach. Numerator degree $3$, denominator degree $2$, toh heuristically yeh $r^3/r^2=r\to 0$ jaisa behave karta hai. Hum L3 mein ek bound se confirm karenge, lekin recognition call ke taur par answer $0$ hai. (Yeh ek forecast hai, abhi proof nahi.)

Numerator degree $4$, denominator degree $2$: $r^4/r^2=r^2\to 0$ jaisa behave karta hai. Forecast: $0$. (L2 mein prove hoga.)

Wohi idea use karo jaise $\frac{x^2}{x^2+y^2}\le 1$. Kyunki $y^2\le x^2+y^2$ hai, isliye $\frac{y^2}{x^2+y^2}\le 1$ hai. Tab $\left|\frac{x^2{y}^2}{x^2+y^2}\right|=x^2\cdot \frac{y^2}{x^2+y^2}\le x^2.$ Jab $(x,y)\to (0,0)$, toh $x^2\to 0$. Squeeze Theorem se limit $\boxed{0}$ hai. Forecast confirm hua.

Polar: $x=r\cos \theta ,y=r\sin \theta$. $\frac{r^3({\cos }^3\theta -{\sin }^3\theta )}{r^2}=r({\cos }^3\theta -{\sin }^3\theta ).$ Polar kyun: denominator mein $r^2$ cleanly cancel ho jaata hai, aur aage $r$ expose ho jaata hai. Kyunki har $\theta$ ke liye $|{\cos }^3\theta -{\sin }^3\theta |\le 2$ hai, hume milta hai $|f|\le 2r\to 0$. Bound $\theta$-independent hai, toh limit $\boxed{0}$ hai.

Polar: numerator $xy=r^2\cos \theta \sin \theta$; denominator $\sqrt{x^2+y^2}=r$. $\frac{r^2\cos \theta \sin \theta }{r}=r\cos \theta \sin \theta .$ Kyunki $|\cos \theta \sin \theta |\le \frac{1}{2}$ hai, hume milta hai $|f|\le \frac{1}{2}r\to 0$. Limit $=\boxed{0}$.

Polar: $\frac{r^2({\cos }^2\theta -{\sin }^2\theta )}{r^2}={\cos }^2\theta -{\sin }^2\theta =\cos 2\theta .$ Yeh kya dikhata hai: value pure $\theta$ hai — koi $r$ nahi bacha. Toh direction $\theta$ ke along andar jaane par aap $\cos 2\theta$ par pahunchte ho. $x$-axis ke along ($\theta =0$): value $1$. $y$-axis ke along ($\theta =\pi /2$): value $-1$. Alag directions → alag values → limit DNE.

Sabhi lines $y=mx$: $\frac{x\cdot m^2{x}^2}{x^2+m^4{x}^4}=\frac{m^{2}x}{1+m^4{x}^2}\to 0.$ Har line $0$ kehti hai. Curve $x=y^2$: $\frac{y^2\cdot y^2}{(y^2{)}^2+y^4}=\frac{y^4}{2y^4}=\frac{1}{2}.$ $x=y^2$ kyun: yeh numerator aur denominator ko same order ($y^4$) banata hai — woh "balanced" path jo lines nahi dekh sakti. Lines $0$ deti hain, curve $\frac{1}{2}$ deta hai → limit DNE. Figure dekho: do paths aur unki level values.

Sabhi lines $y=mx$: $\frac{x^3\cdot mx}{x^6+m^2{x}^2}=\frac{m{x}^4}{x^2(x^4+m^2)}=\frac{m{x}^2}{x^4+m^2}\to \frac{0}{m^2}=0$ ($m\ne 0$ ke liye; $m=0$ bhi $0$ deta hai). Curve $y=x^3$ (orders balance karne ke liye choose kiya — denominator ke dono terms $x^6$ ban jaate hain): $\frac{x^3\cdot x^3}{x^6+(x^3{)}^2}=\frac{x^6}{x^6+x^6}=\frac{1}{2}.$ Lines $0$ deti hain, curve $y=x^3$ pe $\frac{1}{2}$ milta hai → limit DNE.

Continuity ke liye teen cheezein chahiye: $f(0,0)$ exist kare (yeh $0$ hai), limit exist kare, aur dono match karein. Parent Worked Example 3 se, $\left|\frac{x^{2}y}{x^2+y^2}\right|\le |y|\to 0$, toh limit $0$ hai. Kyunki $\lim =0=f(0,0)$, $f$ origin par continuous hai. ✔

Continuity ke liye pehle limit ka exist karna zaroori hai. Lekin $\frac{xy}{x^2+y^2}$ ka polar mein value $\cos \theta \sin \theta$ hai — pure $\theta$, toh $\theta =0$ ke along yeh $0$ hai aur $\theta =\pi /4$ ke along $\frac{1}{2}$ hai. Limit DNE. Kyunki koi single $L$ exist nahi karta, koi bhi $c$ $g$ ko continuous nahi bana sakta. ✗ Key point: continuity ek path-dependent limit ko repair nahi kar sakti — defect surface mein hai, assigned value mein nahi.

$u=x^2+y^2$ substitute karo. Jab $(x,y)\to (0,0)$, toh $u\to 0^{+}$, aur $\frac{\sin u}{u}\to 1$ (standard 1D limit — yahan allowed hai kyunki $h$ sirf $u=r^2$ ke through $(x,y)$ par depend karta hai, toh koi bhi direction matter nahi kar sakta). Isliye limit $1$ hai, aur $h$ origin par continuous hai tabhi jab $h(0,0)=\boxed{1}$ set karein.

Goal: given $\epsilon >0$, ek $\delta >0$ dhundho jisme $0<\sqrt{x^2+y^2}<\delta \Rightarrow \left|\frac{x^{2}y}{x^2+y^2}-0\right|<\epsilon .$ Bound: $\frac{x^2}{x^2+y^2}\le 1$, toh $\left|\frac{x^{2}y}{x^2+y^2}\right|\le |y|\le \sqrt{x^2+y^2}.$ $\delta =\epsilon$ choose karo. Tab $\sqrt{x^2+y^2}<\delta =\epsilon$ force karta hai $|f|<\epsilon$. Yeh poori disk of radius $\delta$ par hold karta hai, sirf paths par nahi — toh yeh airtight hai. $■$

Lo $f(x,y)=\frac{x^2{y}^2}{x^4+y^4}\cdot 0\text{— nahi, deliberately banao:}f(x,y)=\frac{2x^{2}y}{x^4+y^2}.$ Ruko — verify karo. Lines $y=mx$: $\frac{2x^2\cdot mx}{x^4+m^2{x}^2}=\frac{2m{x}^3}{x^2(x^2+m^2)}=\frac{2mx}{x^2+m^2}\to 0.$ Har line $0$ deti hai. ✔ Curve $y=x^2$: $\frac{2x^2\cdot x^2}{x^4+x^4}=\frac{2x^4}{2x^4}=1.$ Is parabola ke along value constantly $1$ hai. ✔ Toh $f$ sabhi lines par $0$ hai phir bhi $y=x^2$ par $1$ tak pahunchta hai — limit DNE hai, aur khaas taur par yeh $0$ nahi hai. Yeh L3·Q2 trap ka exact mechanism hai, value $1$ hit karne ke liye tune kiya gaya. Figure dekho.

$x$-axis ke along ($y=0$): numerator $0$, denominator $(x{)}^2$, toh $f=0\to 0$. $y$-axis ke along ($x=0$): numerator $0$, toh $f=0\to 0$. Diagonal $y=x$ ke along: tab $x-y=0$, toh $f(x,x)=\frac{x^2\cdot x^2}{x^2{x}^2+0}=\frac{x^4}{x^4}=1\to 1.$ Diagonal kyun: term $(x-y{)}^2$ exactly $y=x$ par zero ho jaata hai, denominator ka "spoiler" khatam ho jaata hai aur $1$ bach jaata hai. Axes $0$ dete hain, diagonal $1$ deta hai → limit DNE. $■$