Limits and continuity in 2D — path-dependence issue

WHAT we are asking

HOW path-dependence kills a limit

Standard paths to try, in order (cheapest first):

Along the $x$ -axis: set $y=b$ , let $x\to a$ .
Along the $y$ -axis: set $x=a$ , let $y\to b$ .
Along lines $y = b + m(x-a)$ for general slope $m$ .
Along parabolas like $y=(x-a)^2$ if lines all agree but you suspect trouble.

Worked Example 1 — the canonical failure

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

Path A — along $x$ -axis ( $y=0$ ): $f(x,0)=\frac{x\cdot 0}{x^2+0}=0 \to 0.$ Why this step? Setting $y=0$ collapses the problem to 1D; the numerator vanishes.

Path B — along the line $y=mx$ : $f(x,mx)=\frac{x\cdot mx}{x^2+m^2x^2}=\frac{mx^2}{x^2(1+m^2)}=\frac{m}{1+m^2}.$ Why this step? Substitute $y=mx$ and cancel $x^2$ (legal since $x\neq0$ on the path). The $x$ 's vanish — the value depends only on the slope $m$ .

Verdict: $m=0$ gives $0$ ; $m=1$ gives $\tfrac12$ . Different paths → different values → limit does not exist.

Worked Example 2 — lines lie, parabola tells the truth

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

All straight lines $y=mx$ : $f(x,mx)=\frac{x\cdot m^2x^2}{x^2+m^4x^4}=\frac{m^2x^3}{x^2(1+m^4x^2)}=\frac{m^2x}{1+m^4x^2}\to 0.$ Why this step? Factor $x^2$ from the denominator; the surviving factor $x$ in the numerator forces the limit to $0$ as $x\to0$ . So every line says $0$ .

Now the curve $x=y^2$ : $f(y^2,y)=\frac{y^2\cdot y^2}{(y^2)^2+y^4}=\frac{y^4}{y^4+y^4}=\frac{y^4}{2y^4}=\frac12.$ Why this step? We chose $x=y^2$ to make numerator and denominator the same order ( $y^4$ ). This is the "balanced" path the lines missed.

Verdict: lines give $0$ , parabola gives $\tfrac12$ → limit does not exist. (This is the steel-man trap made concrete.)

Worked Example 3 — limit DOES exist (Squeeze proof)

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

You can't prove this with paths. Use a bound.

Step 1: Note $\dfrac{x^2}{x^2+y^2}\le 1$ for all $(x,y)\neq(0,0)$ . Why? The denominator $x^2+y^2 \ge x^2$ , so the fraction is at most 1.

Step 2: Therefore $\left|\frac{x^2 y}{x^2+y^2}\right| = |y|\cdot\frac{x^2}{x^2+y^2} \le |y|.$ Why this step? Pull out $|y|$ , then bound the leftover fraction by 1.

Step 3: As $(x,y)\to(0,0)$ , $|y|\to0$ . By the Squeeze Theorem the whole thing $\to 0$ . $\boxed{\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^2}=0}$ Why this is airtight: the bound $|f|\le|y|$ holds on the entire disk, so no path can escape it. That's exactly what $\varepsilon$ – $\delta$ needs.

Continuity in 2D

Formula Useful polar-coordinate test

Substitute $x=r\cos\theta,\;y=r\sin\theta$ around $(0,0)$ . Then $(x,y)\to(0,0)\iff r\to0$ . Derivation of the test: if $|f-L|\le g(r)$ where $g(r)\to0$ independently of $\theta$ , the limit is $L$ . If after simplifying $f$ still contains $\theta$ as $r\to0$ , that $\theta$ -dependence is the path-dependence — limit fails.

Example 1 in polar: $\dfrac{xy}{x^2+y^2}=\dfrac{r^2\cos\theta\sin\theta}{r^2}=\cos\theta\sin\theta$ — pure $\theta$ , no $r$ . Depends on direction → no limit. ✔ matches our path result.

Forecast-then-Verify drill

Flashcards

What must be true for a 2D limit to exist?

Every path approaching the point must give the same value (the whole disk, not just two directions).

Can matching finitely many paths prove a 2D limit exists?

No — it can only disprove it (by finding a mismatch). Proof needs

\varepsilon

\delta

or Squeeze.

Limit of

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

at origin?

Does not exist; along

y=mx

it equals

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

, which varies with

m

Why does

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

trap students?

All straight lines give

0

, but the path

x=y^2

gives

\frac12

→ limit DNE.

What does it mean if the polar form keeps a

\theta

r\to0

The value depends on direction → path-dependent → limit does not exist.

State the 3 conditions for continuity of

f

(a,b)

f(a,b)

exists; the limit exists; the limit equals

f(a,b)

Limit of

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

at origin and why?

0

; bound

|f|\le|y|\to0

by Squeeze on the whole disk.

A function homogeneous of degree 0 — what's special about its directional limits?

Constant along each ray, so different rays generally give different values → usually no limit.

Recall Feynman: explain to a 12-year-old

Imagine you and friends all walking toward the same lamppost in a foggy field. In 1D you can only come from the left road or the right road. But in a field you can come from any direction — straight, curvy, zig-zag. A "limit existing" means: no matter which path you walk, the height of the ground right at the lamppost looks the same to everybody. If your friend on the curvy path sees the ground at height 0 but you on the straight path see height ½, then there's a weird crack right at the lamppost — the "limit" is broken. To prove it's smooth you must show everyone, on every imaginable path, ends up at the same height — usually by trapping the height between two squeezing walls that both shrink to the same number.

Connections

Multivariable Calculus — parent chapter
Squeeze Theorem — the tool to prove a 2D limit exists
Polar Coordinates — converts $(x,y)\to(0,0)$ into $r\to0$ for a clean test
Partial Derivatives — built on limits along axis-directions
Differentiability in 2D — requires continuity, which requires path-independent limits
Epsilon-Delta Definition — the 1D ancestor generalized to disks
Continuity in 1D — contrast: only two directions vs. infinitely many

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, 1D limit mein sirf do raaste hote hain — left se aao ya right se. Lekin 2D mein point tak pahunchne ke infinite raaste hote hain: seedhi lines, parabola, curve, spiral, kuch bhi. Limit tabhi exist karti hai jab har ek path same value de. Agar do bhi paths alag-alag answer dein, to limit exist nahi karti — bas yahi pura funda hai (PDD: Paths Disagree → Death).

Classic example: $\frac{xy}{x^2+y^2}$ at origin. $y=0$ pe limit $0$ aata hai, par $y=x$ pe $\frac12$ . Do paths disagree ho gaye → limit DNE. Aur ek bada trap hai $\frac{xy^2}{x^2+y^4}$ — yahan saari straight lines $0$ dengi, par jaise hi tum $x=y^2$ wala curve try karoge, $\frac12$ aa jaata hai. Isliye yaad rakho: "Lines Lie, Curves Confess" — sirf lines match ho jaane se khush mat ho jao.

Important baat: paths se tum sirf limit ko galat saabit kar sakte ho (mismatch dhoondh ke). Limit exist karti hai ye prove karne ke liye paths kaafi nahi — Squeeze Theorem ya $\varepsilon$ - $\delta$ bound chahiye. Jaise $\frac{x^2y}{x^2+y^2}$ mein hum dikhate hain $|f|\le|y|\to0$ , aur ye bound poore disk pe valid hai, isliye limit $0$ hai. Ek aur tagda trick: polar coordinates. $x=r\cos\theta, y=r\sin\theta$ daal do; agar $r\to0$ ke baad bhi $\theta$ bacha hua hai, matlab direction pe depend kar raha hai → limit fail. Continuity matlab limit exist kare aur $f(a,b)$ ke barabar ho — agar limit hi path-dependent hai to function kabhi continuous nahi ban sakta.

Limits and continuity in 2D — path-dependence issue

WHAT we are asking

HOW path-dependence kills a limit

Worked Example 1 — the canonical failure

Worked Example 2 — lines lie, parabola tells the truth

Worked Example 3 — limit DOES exist (Squeeze proof)

Continuity in 2D

Forecast-then-Verify drill

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Multivariable Calculus

Connections