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

Along $x$-axis set $y=0$: $f(x,0)=\frac{x-0}{x+0}=1\to 1.$ Along $y$-axis set $x=0$: $f(0,y)=\frac{0-y}{0+y}=-1\to -1.$ WHAT we did: collapsed to 1D twice. WHY: cheapest paths first. They give $1$ and $-1$. Verdict: two paths disagree $\Rightarrow$ limit DNE.

True. Numerator degree $3$, denominator degree $2$, so heuristically it behaves like $r^3/r^2=r\to 0$. We confirm with a bound in L3, but as a recognition call the answer is $0$. (This is a forecast, not yet a proof.)

Numerator degree $4$, denominator degree $2$: behaves like $r^4/r^2=r^2\to 0$. Forecast: $0$. (Proved in L2.)

Use the same idea as $\frac{x^2}{x^2+y^2}\le 1$. Since $y^2\le x^2+y^2$, we have $\frac{y^2}{x^2+y^2}\le 1$. Then $\left|\frac{x^2{y}^2}{x^2+y^2}\right|=x^2\cdot \frac{y^2}{x^2+y^2}\le x^2.$ As $(x,y)\to (0,0)$, $x^2\to 0$. By the Squeeze Theorem the limit is $\boxed{0}$. Forecast confirmed.

Polar: $x=r\cos \theta ,y=r\sin \theta$. $\frac{r^3({\cos }^3\theta -{\sin }^3\theta )}{r^2}=r({\cos }^3\theta -{\sin }^3\theta ).$ WHY polar: the $r^2$ in the denominator cancels cleanly, exposing an $r$ in front. Since $|{\cos }^3\theta -{\sin }^3\theta |\le 2$ for every $\theta$, we get $|f|\le 2r\to 0$. The bound is $\theta$-independent, so the limit is $\boxed{0}$.

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

Polar: $\frac{r^2({\cos }^2\theta -{\sin }^2\theta )}{r^2}={\cos }^2\theta -{\sin }^2\theta =\cos 2\theta .$ WHAT this shows: the value is pure $\theta$ — no $r$ left. So walking in along direction $\theta$ you land on $\cos 2\theta$. Along $x$-axis ($\theta =0$): value $1$. Along $y$-axis ($\theta =\pi /2$): value $-1$. Different directions → different values → limit DNE.

All lines $y=mx$: $\frac{x\cdot m^2{x}^2}{x^2+m^4{x}^4}=\frac{m^{2}x}{1+m^4{x}^2}\to 0.$ Every line says $0$. Curve $x=y^2$: $\frac{y^2\cdot y^2}{(y^2{)}^2+y^4}=\frac{y^4}{2y^4}=\frac{1}{2}.$ WHY $x=y^2$: it makes numerator and denominator the same order ($y^4$) — the "balanced" path lines cannot see. Lines give $0$, curve gives $\frac{1}{2}$ → limit DNE. See figure: the two paths and their level values.

All lines $y=mx$: $\frac{x^3\cdot mx}{x^6+m^2{x}^2}=\frac{m{x}^4}{x^2(x^4+m^2)}=\frac{m{x}^2}{x^4+m^2}\to \frac{0}{m^2}=0$ (for $m\ne 0$; $m=0$ gives $0$ too). Curve $y=x^3$ (chosen to balance orders — denominator terms both become $x^6$): $\frac{x^3\cdot x^3}{x^6+(x^3{)}^2}=\frac{x^6}{x^6+x^6}=\frac{1}{2}.$ Lines give $0$, curve $y=x^3$ gives $\frac{1}{2}$ → limit DNE.

Continuity needs three things: $f(0,0)$ exists (it's $0$), the limit exists, and they match. From the parent Worked Example 3, $\left|\frac{x^{2}y}{x^2+y^2}\right|\le |y|\to 0$, so the limit is $0$. Since $\lim =0=f(0,0)$, $f$ is continuous at the origin. ✔

Continuity requires the limit to exist first. But $\frac{xy}{x^2+y^2}$ has, in polar, value $\cos \theta \sin \theta$ — pure $\theta$, so along $\theta =0$ it is $0$ and along $\theta =\pi /4$ it is $\frac{1}{2}$. Limit DNE. Since no single $L$ exists, no choice of $c$ can make $g$ continuous. ✗ Key point: continuity cannot repair a path-dependent limit — the defect is in the surface, not in the assigned value.

Substitute $u=x^2+y^2$. As $(x,y)\to (0,0)$, $u\to 0^{+}$, and $\frac{\sin u}{u}\to 1$ (the standard 1D limit — allowed here because $h$ depends on $(x,y)$ only through $u=r^2$, so no direction can matter). Thus the limit is $1$, and $h$ is continuous at the origin iff we set $h(0,0)=\boxed{1}$.

Goal: given $\epsilon >0$, find $\delta >0$ with $0<\sqrt{x^2+y^2}<\delta \Rightarrow \left|\frac{x^{2}y}{x^2+y^2}-0\right|<\epsilon .$ Bound: $\frac{x^2}{x^2+y^2}\le 1$, so $\left|\frac{x^{2}y}{x^2+y^2}\right|\le |y|\le \sqrt{x^2+y^2}.$ Choose $\delta =\epsilon$. Then $\sqrt{x^2+y^2}<\delta =\epsilon$ forces $|f|<\epsilon$. This holds on the whole disk of radius $\delta$, not merely on paths — so it is airtight. $■$

Take $f(x,y)=\frac{x^2{y}^2}{x^4+y^4}\cdot 0\text{— no, build it deliberately:}f(x,y)=\frac{2x^{2}y}{x^4+y^2}.$ Wait — verify. Lines $y=mx$: $\frac{2x^2\cdot mx}{x^4+m^2{x}^2}=\frac{2m{x}^3}{x^2(x^2+m^2)}=\frac{2mx}{x^2+m^2}\to 0.$ Every line gives $0$. ✔ Curve $y=x^2$: $\frac{2x^2\cdot x^2}{x^4+x^4}=\frac{2x^4}{2x^4}=1.$ Along this parabola the value is constantly $1$. ✔ So $f$ is $0$ on all lines yet reaches $1$ on $y=x^2$ — the limit DNE, and in particular it is not $0$. This is the exact mechanism of the L3·Q2 trap, tuned to hit value $1$. See figure.

Along $x$-axis ($y=0$): numerator $0$, denominator $(x{)}^2$, so $f=0\to 0$. Along $y$-axis ($x=0$): numerator $0$, so $f=0\to 0$. Along the diagonal $y=x$: then $x-y=0$, so $f(x,x)=\frac{x^2\cdot x^2}{x^2{x}^2+0}=\frac{x^4}{x^4}=1\to 1.$ WHY the diagonal: the term $(x-y{)}^2$ vanishes exactly on $y=x$, killing the "spoiler" in the denominator and leaving $1$. Axes give $0$, diagonal gives $1$ → limit DNE. $■$