For a function of one variable, approaching a point means coming in from the left or from the
right — only two directions. A
function of two variables
lives over the plane, and a point in the plane can be approached along infinitely many
paths: straight lines at every angle, parabolas, spirals. The limit
\lim_{(x, y) \to (a, b)} f(x, y) = L
means f(x, y) gets arbitrarily close to L
as (x, y) gets close to (a, b) —
no matter how it gets there. That last clause is the whole subtlety of the
subject. If two different routes into (a, b) disagree on the
approached value, the limit simply does not exist.
The cautionary tale, step by step
The standard warning example is
f(x, y) = \frac{xy}{x^2 + y^2},
which is defined everywhere except the origin. We ask: does
\lim_{(x, y) \to (0, 0)} f(x, y) exist? We test two straight-line
approaches.
Step 1 — approach along the x-axis (y = 0).
Substitute y = 0 and let x \to 0:
f(x, 0) = \frac{x \cdot 0}{x^2 + 0^2} = \frac{0}{x^2} = 0 \quad\Longrightarrow\quad \lim_{x \to 0} f(x, 0) = 0.
Step 2 — approach along the diagonal (y = x).
Substitute y = x and let x \to 0:
f(x, x) = \frac{x \cdot x}{x^2 + x^2} = \frac{x^2}{2x^2} = \frac{1}{2} \quad\Longrightarrow\quad \lim_{x \to 0} f(x, x) = \frac{1}{2}.
Step 3 — compare the two routes. Along the axis the function heads for
0; along the diagonal it sits constantly at
\tfrac12. The destination depends on the road taken:
0 = \lim_{x \to 0} f(x, 0) \;\ne\; \lim_{x \to 0} f(x, x) = \frac{1}{2}.
Step 4 — conclude. Because two paths give two different values, no single
number L can be the limit along every path. Hence
\lim_{(x, y) \to (0, 0)} \frac{xy}{x^2 + y^2} \ \text{does not exist.}
Notice the trap: each individual path-limit exists and is finite. The failure is in
their disagreement. Finding one path that misbehaves is enough to kill a
limit; finding several that agree proves nothing — there is always another path you
did not try. Drag the approach line below to see the value on the diagonal pull away from the
value on the axis.
Continuity
Once limits are pinned down, continuity is the familiar slogan: f
is continuous at (a, b) when the limit exists and
equals the value there,
\lim_{(x, y) \to (a, b)} f(x, y) = f(a, b).
Sums, products, quotients (away from zero denominators) and compositions of continuous
functions are continuous, so polynomials like x^2 + y^2 are
continuous everywhere. The troublemaker above is continuous on all of
\mathbb{R}^2 except the origin — and no choice of
f(0, 0) can repair it, precisely because the limit there does not
exist.
For f : D \subseteq \mathbb{R}^2 \to \mathbb{R} and a point
(a, b):
-
If \lim_{(x, y) \to (a, b)} f(x, y) = L exists, then
f \to L along every path approaching
(a, b) — all path-limits agree on L.
-
Two-path test (contrapositive). If two paths give two different
path-limits — or even one path-limit fails to exist — then the limit
does not exist.
-
Agreement along finitely many paths is never sufficient to prove a limit
exists; existence requires a genuine bound holding uniformly in all directions (e.g. via
polar coordinates or a squeeze).
The two-path test can only disprove a limit. To prove one exists you must control
every direction at once, and the cleanest tool is the
polar substitution
x = r\cos\theta, y = r\sin\theta, where
r = \sqrt{x^2 + y^2} is the distance to the origin. Approaching
(0,0) is exactly r \to 0, for
any angle \theta. Consider
g(x, y) = \frac{x^2 y}{x^2 + y^2}.
In polar form the denominator collapses, x^2 + y^2 = r^2, and
g = \frac{(r\cos\theta)^2 (r\sin\theta)}{r^2} = r\,\cos^2\theta\,\sin\theta.
Now |\cos^2\theta\,\sin\theta| \le 1 for every angle, so
|g| \le r regardless of \theta. As
r \to 0 the bound forces g \to 0
along every path simultaneously — the angle dropped out of the bound, which is the whole
point. Hence \lim_{(x, y) \to (0, 0)} g = 0 genuinely exists.
(Contrast the bad example: there the polar form is
\cos\theta\sin\theta = \tfrac12\sin 2\theta, with no
r out front to crush it — the angle survives, so the value
depends on direction.)