Limits and continuity in 2D — path-dependence issue: Question bank

Before we start, one reminder of the vocabulary so nothing here is unearned:

A path is any curve (line, parabola, spiral) that ends at the target point $(a, b)$ .
The limit exists only if every path gives the same value $L$ — because the $ε$ – $δ$ definition demands a whole disk works, not a few roads.
Matching paths can only DISprove a limit (find a mismatch). Proving existence needs the Squeeze Theorem or an $ε$ – $δ$ bound.

True or false — justify

If two straight-line paths through the origin give the same value, the limit exists.

False. Two — or even infinitely many lines — can agree by coincidence;

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

gives

0

on every line yet along

x = y^{2}

reduces to

\frac{y ^{2} \cdot y ^{2}}{( y ^{2} ) ^{2} + y ^{4}} = \frac{y ^{4}}{y ^{4} + y ^{4}} = \frac{y ^{4}}{2 y ^{4}} = \frac{1}{2}

. Agreement only rules limits out, never in.

If a limit exists along every straight line, then it exists.

False. "Every line" still misses curved paths like parabolas and spirals; the balanced curve

x = y^{2}

is the classic escapee. Lines lie, curves confess.

If one path gives

0

and another gives

\frac{1}{2}

, the limit does not exist.

True. A single genuine disagreement between two paths violates "all paths agree," so no common

L

can exist. This is the whole power of the killer test.

A function homogeneous of degree

0

can still have a limit at the origin.

Rarely, but yes: only if it is constant on all rays (same value every direction). If any two rays differ, its directional constancy guarantees no limit. Usually it fails.

If the polar form of

f

still contains

θ

r \to 0

, the limit does not exist.

True precisely when the

θ

-terms are not killed by a factor

\to 0

. The clean criterion: write

f = g (r, θ)

; if you can bound

∣ f - L ∣ \leq h (r)

with

h (r) \to 0

independent of $θ$ , the limit is

L

. If instead the

θ

-dependence survives as

r \to 0

(e.g.

cos θ sin θ

, a bare function of

θ

with no shrinking

r

-factor), the limit does not exist. So

r cos θ

is harmless (

r \to 0

crushes it); a lone

cos θ sin θ

is fatal.

Continuity at

(a, b)

only requires the limit to exist there.

False. It needs three things:

f (a, b)

exists, the limit exists, and they are equal. A hole or a jump at the point breaks continuity even if a limit exists.

lim_{(x, y) \to (0, 0)} f = L

exists, then approaching along

y = x^{2}

also gives

L

True. Existence means every path gives

L

, so any specific path — including a parabola — must agree. This is why one deviating path is fatal.

∣ f (x, y) ∣ \leq ∣ y ∣

near the origin is enough to conclude

lim f = 0

True, provided the bound holds on the whole disk: as

∣ y ∣ \to 0

the Squeeze Theorem forces

f \to 0

on every path at once, which is exactly what

ε

–

δ

needs.

Spot the error

"I checked

y = 0

and

x = 0

; both gave

0

, so the limit is

0

The error is claiming existence from two paths. Two axes are just two directions out of infinitely many; they can only support a guess, never a proof — try slopes and a parabola, then Squeeze.

"Along

y = m x

(with

m \in R

the slope),

\frac{x y}{x ^{2} + y ^{2}} = \frac{m}{1 + m ^{2}}

; plugging

m = 0

gives

0

, so limit is

0

The value

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

depends on $m$ — that dependence itself is the proof the limit fails (

m = 1

gives

\frac{1}{2}

). Reading off a single

m

hides the disagreement.

"In polar,

\frac{x y}{x ^{2} + y ^{2}} = cos θ sin θ \to 0

r \to 0

There is no

r

to send to

0

; the expression is pure $θ$ . Its value stays

cos θ sin θ

forever, so it never settles — the limit does not exist.

f (y^{2}, y) = \frac{y ^{4}}{2 y ^{4}} = \frac{1}{2}

, and

\frac{1}{2} \neq = 0

, so

f

is discontinuous — but only on that one curve."

Discontinuity is a property at the point, not "on a curve." The mismatch means no value

L

exists at the origin, so

f

cannot be made continuous there by any assignment.

"The limit exists because I can bound

∣ f ∣ \leq r^{2} cos θ

A valid bound must not depend on direction;

r^{2} cos θ

still carries

θ

. Bound instead by

r^{2}

alone (since

∣ cos θ ∣ \leq 1

) so it shrinks uniformly to

0

"Since each partial derivative exists at

(0, 0)

f

is continuous there."

Partial Derivatives only probe the two axis directions; their existence says nothing about curved approaches. A function can have both partials yet fail continuity (path-dependent limit).

Why questions

Why must the

ε

–

δ

condition use a disk and not a line?

The disk forces the output near

L

from all directions simultaneously, encoding "every path agrees." A line would only control one approach and miss path-dependence entirely.

Why can matching paths never prove a limit exists?

You can only ever test a countable handful of the infinitely many paths; the untested ones (like the balanced curve) may disagree. Absence of a found mismatch is not proof of none.

Why does converting to Polar Coordinates help at the origin?

It turns "

(x, y) \to (0, 0)

" into the single condition "

r \to 0

," separating how far (

r

) from which direction (

θ

). If

θ

then vanishes from the bound, all directions collapse to one value.

Why is

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

constant along each ray from the origin?

It is homogeneous of degree

0

: scaling

(x, y) \to (t x, t y)

leaves it unchanged, so its value depends only on the ray's direction, not on distance. Different rays therefore lock in different constants.

Why do we try the parabola

x = y^{2}

specifically when all lines agree?

It makes numerator and denominator the same order (in

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

both become

y^{4}

), so the offending term no longer vanishes — exactly the "balanced" path lines cannot reach.

Why is the Squeeze Theorem the right tool to prove a 2D limit exists?

It replaces "check every path" with a single inequality

∣ f - L ∣ \leq g (r) \to 0

holding on the whole disk, which certifies all paths at once — matching the

ε

–

δ

requirement directly.

Why does the naive line test

y = m x

automatically fail to catch the parabola trap?

Lines force

y

proportional to

x

, so

y^{4} = m^{4} x^{4}

stays a higher order than

x^{2}

and dies; only when

x

scales like

y^{2}

do the orders match and the hidden value appears.

Edge cases

What happens to the limit if

f

is undefined only at

(0, 0)

but the limit exists?

The function has a removable discontinuity: define

f (0, 0) = L

and it becomes continuous. Existence of the limit is precisely what makes the repair possible.

Approaching

(0, 0)

along the constant path "stay at the origin" — is that allowed?

No. The definition requires

0 < x^{2} + y^{2}

, i.e. strictly away from the point; the value at the point never enters the limit, only the surrounding disk does.

f \equiv 3

everywhere except a single ray where

f = 5

, does the limit at the origin exist?

No. Along that ray the approach gives

5

while every other direction gives

3

; one deviating direction, no matter how "thin," breaks agreement and kills the limit.

Can a limit exist along every parabola

y = c x^{2}

yet still fail?

Yes. Just as lines miss parabolas, parabolas can miss other curves (e.g.

y = x^{3}

or spirals). No finite family of paths is ever complete — only a uniform bound settles it.

What if the polar bound is

∣ f ∣ \leq r

but only for

θ \neq = \frac{π}{2}

The bound is not uniform: it fails on one direction, so you have not controlled the whole disk. You must handle the excluded direction separately or the "proof" has a hole.

Along a spiral

r = θ

into the origin, does

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

have a limit?

No. On the spiral

θ

keeps changing without settling, so

cos θ sin θ

oscillates all the way in — the spiral samples many directions and never converges.

If both one-sided 1D limits of a restriction match, is the 2D limit safe?

No. A restriction is a single path; its two "sides" are still one direction of approach in 2D. This borrows 1D confidence that simply does not transfer to the plane.

Connections

Multivariable Calculus — parent chapter
Squeeze Theorem — the only reliable way to prove existence
Polar Coordinates — separates distance $r$ from direction $θ$
Partial Derivatives — axis-only limits, insufficient for continuity
Differentiability in 2D — demands continuity, hence path-independence
Epsilon-Delta Definition — the disk condition behind everything
Continuity in 1D — where the misleading two-sided intuition comes from