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

This page is a drill floor. The parent note built the idea; here we hit every kind of case the topic can throw at you, one worked example per cell. Before each solution there is a Forecast line — cover the answer and guess first. That habit is the whole game.

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The scenario matrix

Every 2D-limit question you will meet lives in exactly one of these boxes. Our examples below cover all of them.

Cell	What makes it that cell	Killer tool	Example
A. Line test kills it	Two straight lines already disagree	Two paths	Ex 1
B. Lines lie, curve confesses	All lines agree, a parabola/cubic breaks it	Smart curve	Ex 2
C. Genuinely exists (numerator wins)	Top degree beats bottom → shrinks	Squeeze / polar	Ex 3
D. Sign / quadrant sensitivity	Value flips depending on which quadrant you enter	Signed paths	Ex 4
E. Degenerate / removable-looking	Looks like $0/0$ but is actually harmless	Algebra + polar	Ex 5
F. Non-origin target point	Approach a point $(a,b)\ne (0,0)$	Shift coordinates	Ex 6
G. Real-world word problem	Physical quantity, must interpret DNE	Model + paths	Ex 7
H. Exam twist: choose $k$ so continuous	Solve for the value that patches the hole	Limit = value	Ex 8

The columns "signs/quadrants", "zero/degenerate inputs", "limiting values", "word problem" and "exam twist" from the task are Cells D, E, C, G, H respectively. Let's walk them.

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Cell A — the line test alone finishes it

Forecast: Top and bottom are both degree 2 — same "size." When top and bottom scale together, the $x$'s and $y$'s usually don't cancel, so I bet the answer depends on direction → DNE.

Step 1 — Approach along the $x$-axis ($y=0$). $f(x,0)=\frac{x^2-0}{x^2+0}=1\to 1.$ Why this step? Setting $y=0$ is the cheapest path — it collapses the two-variable mess to one variable. The $x$-axis gives $1$.

Step 2 — Approach along the $y$-axis ($x=0$). $f(0,y)=\frac{0-y^2}{0+y^2}=-1\to -1.$ Why this step? The other free axis is the next cheapest path. It already gives $-1$.

Step 3 — Compare. Path along $x$-axis says $1$; path along $y$-axis says $-1$. Two paths, two answers.

Verify (polar sanity check): with $x=r\cos \theta ,y=r\sin \theta$, $\frac{r^2{\cos }^2\theta -r^2{\sin }^2\theta }{r^2}={\cos }^2\theta -{\sin }^2\theta =\cos 2\theta .$ The $r$ cancelled completely — pure direction $\theta$ remains. $\theta =0$ gives $1$, $\theta =\frac{\pi }{2}$ gives $-1$. Confirmed: limit DNE.

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Cell B — lines lie, curve confesses

Forecast: This is a cousin of the parent's $\frac{x{y}^2}{x^2+y^4}$ trap. I predict every straight line gives $0$, but a well-chosen parabola breaks it.

Step 1 — Every straight line $y=mx$. $f(x,mx)=\frac{x^2\cdot mx}{x^4+m^2{x}^2}=\frac{m{x}^3}{x^2(x^2+m^2)}=\frac{mx}{x^2+m^2}\underset{x\to 0}{\to }\frac{0}{m^2}=0.$ Why this step? Factor $x^2$ from the denominator; a lone $x$ survives on top and drags the whole thing to $0$. Every line says $0$ (for $m\ne 0$; the axes give $0$ too by inspection).

Step 2 — Try the balanced curve $y=x^2$. We want the numerator and denominator to be the same order. Numerator $x^{2}y$ becomes $x^2\cdot x^2=x^4$; denominator becomes $x^4+x^4$. Match! $f(x,x^2)=\frac{x^2\cdot x^2}{x^4+(x^2{)}^2}=\frac{x^4}{x^4+x^4}=\frac{x^4}{2x^4}=\frac{1}{2}.$ Why this step? The lines missed the direction where top and bottom shrink at the same rate. That's always the curve to hunt for.

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

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Cell C — it genuinely exists (limiting value)

Forecast: Numerator is degree 4, denominator degree 2. Like $r^4/r^2=r^2\to 0$. I bet the limit is $0$ — and paths can't prove it, so I'll need Squeeze or polar.

Step 1 — Convert to polar. Set $x=r\cos \theta ,y=r\sin \theta$. $\frac{r^4{\cos }^4\theta +r^4{\sin }^4\theta }{r^2}=r^2({\cos }^4\theta +{\sin }^4\theta ).$ Why this step? Polar separates "how close" ($r$) from "which direction" ($\theta$). If the answer still has $\theta$ as $r\to 0$, it's path-dependent; if $\theta$ only sits inside a bounded factor, we can squeeze it.

Step 2 — Bound the direction factor. Since ${\cos }^4\theta \le 1$ and ${\sin }^4\theta \le 1$, $0\le {\cos }^4\theta +{\sin }^4\theta \le 2.$ Why this step? We don't need the exact value of the trig part — only that it can't blow up. Bounding it kills all direction-dependence.

Step 3 — Squeeze. $0\le \left|\frac{x^4+y^4}{x^2+y^2}\right|=r^2({\cos }^4\theta +{\sin }^4\theta )\le 2r^2.$ As $(x,y)\to (0,0)$, $r\to 0$, so $2r^2\to 0$. Both walls collapse to $0$. By the Squeeze Theorem the middle is forced to $0$ on the whole disk — no path escapes. $\boxed{\underset{(x,y)\to (0,0)}{\lim }\frac{x^4+y^4}{x^2+y^2}=0.}$

Verify: try any line $y=mx$: $\frac{x^4+m^4{x}^4}{x^2+m^2{x}^2}=\frac{x^2(1+m^4)}{1+m^2}\to 0$. Consistent with the boxed answer.

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Cell D — sign / quadrant sensitivity

Forecast: The $|x|$ (absolute value of $x$) means the sign of $x$ is wiped out but the sign of $y$ survives. Along $y=mx$ I expect the value to still depend on the slope's sign → DNE, and the $|x|$ should make quadrants I & II behave alike but flip against III & IV.

Step 1 — Line $y=mx$ with $x>0$ (so $|x|=x$). $f(x,mx)=\frac{x\cdot mx}{x^2+m^2{x}^2}=\frac{m{x}^2}{x^2(1+m^2)}=\frac{m}{1+m^2}.$ Why this step? Fix $x>0$ so $|x|$ is just $x$; then it's the parent's Example-1 algebra. Slope $m$ survives → already directional.

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

Step 3 — See the quadrant story (why $|x|$ matters). Enter from the left, $x<0$, along $y=mx$ so now $|x|=-x$: $f(x,mx)=\frac{(-x)(mx)}{x^2(1+m^2)}=\frac{-m{x}^2}{x^2(1+m^2)}=\frac{-m}{1+m^2}.$ Why this step? With $x<0$, the $|x|$ flips sign of the numerator. So approaching with slope $m$ from the right gives $\frac{m}{1+m^2}$ but from the left gives $\frac{-m}{1+m^2}$ — the two half-lines of the same geometric line disagree! This is pure quadrant sensitivity.

Verify (polar): $|x|=r|\cos \theta |$, so $\frac{r|\cos \theta |\cdot r\sin \theta }{r^2}=|\cos \theta |\sin \theta ,$ pure $\theta$. At $\theta =\frac{\pi }{4}$: $\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=\frac{1}{2}$. At $\theta =\frac{3\pi }{4}$ (quadrant II): $|-\frac{1}{\sqrt{2}}|\cdot \frac{1}{\sqrt{2}}=\frac{1}{2}$ — same as I. At $\theta =-\frac{\pi }{4}$ (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 but harmless

Forecast: At $(0,0)$ this is $\frac{0}{0}$ — the classic "degenerate" indeterminate form that scares people. But numerator degree 4 beats denominator degree 2, so I expect it to vanish: limit $=0$.

Step 1 — Bound one factor. Notice $\frac{y^2}{x^2+y^2}\le 1$ because the denominator is at least $y^2$. Why this step? We want to peel off a piece that provably shrinks and cap the leftover by $1$.

Step 2 — Rewrite and bound. $0\le \frac{x^2{y}^2}{x^2+y^2}=x^2\cdot \frac{y^2}{x^2+y^2}\le x^2.$ Why this step? Pull out $x^2$ (which clearly $\to 0$) and cap the remaining fraction by $1$. The messy denominator is now tamed.

Step 3 — Squeeze. As $(x,y)\to (0,0)$, $x^2\to 0$, so the middle is trapped to $0$. $\boxed{\underset{(x,y)\to (0,0)}{\lim }\frac{x^2{y}^2}{x^2+y^2}=0.}$

Verify (polar): $\frac{r^2{\cos }^2\theta r^2{\sin }^2\theta }{r^2}=r^2{\cos }^2\theta {\sin }^2\theta$, and $|{\cos }^2\theta {\sin }^2\theta |\le \frac{1}{4}$, so $|f|\le \frac{1}{4}{r}^2\to 0$. The "$0/0$" was never a real problem — the top just wins.

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Cell F — target point that is not the origin

Forecast: This is Example-1's DNE structure ($\frac{uv}{u^2+v^2}$) but centred at $(1,2)$ instead of the origin. I bet a coordinate shift turns it into the exact same "depends on slope" failure → DNE.

Step 1 — Shift coordinates. Let $u=x-1,v=y-2$. Why this step? The $\epsilon$–$\delta$ definition only cares about distance from the target. Sliding the whole plane so $(1,2)$ lands on the origin changes nothing about the limit but makes the algebra familiar. As $(x,y)\to (1,2)$, we get $(u,v)\to (0,0)$.

Step 2 — Rewrite. $\frac{(x-1)(y-2)}{(x-1{)}^2+(y-2{)}^2}=\frac{uv}{u^2+v^2}.$ Why this step? Identical to parent Example 1. We already proved this is directional.

Step 3 — Reuse the known result. Along $v=mu$: value $=\frac{m}{1+m^2}$, which changes with $m$ ($m=0\to 0$, $m=1\to \frac{1}{2}$).

Verify: the limit DNE at $(1,2)$ — the hole is at the shifted origin, exactly as expected. (Sanity: plug the path $y=2$, i.e. $v=0$: $\frac{0}{u^2}=0$; plug $y-2=x-1$, i.e. $v=u$: $\frac{u^2}{2u^2}=\frac{1}{2}$. Mismatch confirmed.)

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Cell G — real-world word problem

Forecast: This is Example 1 wearing a physics costume. If the mathematical limit DNE, then physically there is no consistent temperature to fill the hole — the plate has a genuine "crack" in its reading at the centre.

Step 1 — Restate as a limit. "Smooth reading at the origin" means $\underset{(x,y)\to (0,0)}{\lim }T(x,y)$ must exist and we set $T(0,0)$ equal to it (the continuity conditions from the parent note). Why this step? Continuity = limit exists and equals the assigned value. First we must know the limit even exists.

Step 2 — Walk two physical routes to the centre. Approach the origin along the wire $y=0$ (a straight scratch across the plate): $T(x,0)=\frac{0}{x^2}=0^{\circ }$C. Approach along the diagonal $y=x$: $T(x,x)=\frac{x^2}{2x^2}=0.5^{\circ }$C. Why this step? Two physical paths, two different temperatures approaching the very same point.

Step 3 — Interpret. Because a sensor sliding in along the scratch would read $0^{\circ }$ but one sliding in diagonally would read $0.5^{\circ }$, no single value patches the hole. The limit DNE, so the plate is unavoidably discontinuous at the centre.

Verify (units + polar): $T=\cos \theta \sin \theta =\frac{1}{2}\sin 2\theta$ °C — a dimensionless ratio times °C, units check out. Its value ranges over $[-\frac{1}{2},\frac{1}{2}]$ °C purely by direction; e.g. $\theta =\frac{\pi }{4}\Rightarrow 0.5$°C, $\theta =0\Rightarrow 0$°C. No consistent origin temperature exists.

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Cell H — exam twist: pick $k$ to make it continuous

Forecast: The parent's Forecast-then-Verify drill already handled this fraction: degree 3 over degree 2 behaves like $r\to 0$, so the limit is $0$. If a limit exists, the patching value must equal it — so I predict $k=0$.

Step 1 — Compute the limit (polar). $\frac{r^3({\cos }^3\theta +{\sin }^3\theta )}{r^2}=r({\cos }^3\theta +{\sin }^3\theta ).$ Why this step? Continuity demands the limit; polar gives it cleanly and exposes any direction-dependence.

Step 2 — Bound and squeeze. Since $|{\cos }^3\theta +{\sin }^3\theta |\le 2$, $0\le |f|\le 2r\underset{r\to 0}{\to }0.$ So $\underset{(x,y)\to (0,0)}{\lim }f=0$ — direction-independent, limit exists. Why this step? The bounded trig factor means no path disagrees; the limit is genuinely $0$.

Step 3 — Match value to limit. Continuity needs $f(0,0)=\lim f$. Therefore $\boxed{k=0.}$ Any other $k$ leaves a jump: the surrounding values approach $0$ but the assigned centre would sit at $k\ne 0$ — a discontinuity.

Verify: with $k=0$, all three continuity conditions hold — $f(0,0)=0$ exists, the limit $=0$ exists, and they're equal. Line check $y=x$: $\frac{2x^3}{2x^2}=x\to 0$. ✔

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Case-coverage recap

Recall Did we hit every cell?

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 — $\frac{|x|y}{x^2+y^2}$ flips between upper/lower half. 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 — shift to origin, DNE at $(1,2)$. Cell G (word problem) ::: Ex 7 — heated plate, no patchable centre temperature. Cell H (choose $k$) ::: Ex 8 — $k=0$ makes it continuous.

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Connections

• Multivariable Calculus — parent chapter
• Parent topic note
• Squeeze Theorem — proves Cells C, E, H
• Polar Coordinates — the $r,\theta$ split behind every "exists?" verdict
• Epsilon-Delta Definition — why the whole disk matters, not just paths
• Continuity in 1D — the two-sided intuition we generalise here
• Partial Derivatives — the axis-direction limits reappear there
• Differentiability in 2D — needs the continuity we test on this page