Limits and Continuity in Several Variables

Does a function of two variables settle down to a single value? This must hold as we approach a point from every possible direction at once. This question is subtler than it looks. In single-variable calculus, a point on the line can only be approached from the left or the right. So checking a limit means checking two directions. In Rn\mathbb{R}^n for n≥2n \geq 2, a point can be approached along infinitely many paths. These include straight lines of every slope, parabolas, spirals, and sequences that zig-zag erratically. The limit exists only if the function converges to the same value along all of them simultaneously. This distinction is not a technicality. It is the central new phenomenon of multivariable calculus. Functions that look perfectly innocent, built from ratios of polynomials, can fail to have a limit at the origin. This happens precisely because different paths of approach disagree. Understanding this failure mode is essential. So is building a rigorous apparatus to detect and rule it out. This is the foundation on which differentiability, the chain rule, and all of vector calculus are built. The payoff for getting this right is large. Continuity in several variables is what makes optimization problems well-posed. Without it, the Extreme Value Theorem fails. "Find the maximum of ff on this closed bounded region" becomes meaningless. Continuity is also what allows us to compute limits of vector-valued functions componentwise. This reduces an Rn→Rm\mathbb{R}^n \to \mathbb{R}^m problem to mm ordinary scalar problems. But first we must establish that this reduction is valid. This article gives the precise ε\varepsilon–δ\delta definition of a multivariable limit. It explains why "path tests" only ever disprove limits and never prove them. It proves the componentwise criterion for vector-valued functions. It also proves the Extreme Value Theorem for continuous functions on compact subsets of Rn\mathbb{R}^n. Throughout, we build on Euclidean Space and Norms and Open and Closed Sets in R^n. We also set the stage for Partial Derivatives and Differentiability and the Total Derivative.

Key takeaways
  • A limit lim⁡x→af(x)=L\lim_{x\to a} f(x) = L in Rn\mathbb{R}^n means: for every ε>0\varepsilon>0 there is δ>0\delta>0 such that 0<∥x−a∥<δ  ⟹  ∥f(x)−L∥<ε0<\|x-a\|<\delta \implies \|f(x)-L\|<\varepsilon — one δ\delta must work for all directions.
  • Approaching along different paths (lines, parabolas, curves) that give different limiting values proves the limit does not exist; agreement along finitely many paths never proves it does.
  • Continuity at aa means ff is defined at aa and lim⁡x→af(x)=f(a)\lim_{x\to a} f(x) = f(a); it is equivalent to the topological "preimage of open is open" definition.
  • A vector-valued function f=(f1,…,fm):Rn→Rmf = (f_1,\dots,f_m): \mathbb{R}^n \to \mathbb{R}^m is continuous iff each real-valued component fif_i is continuous — this is the componentwise criterion.
  • The Extreme Value Theorem: a continuous real-valued function on a nonempty compact subset of Rn\mathbb{R}^n attains a global maximum and minimum.
  • Key idea: in several variables, "the limit exists" is a statement about uniform agreement over an entire (n-1)-sphere of directions at once, not a finite check of cases.

Where this lives today

Multivariable continuity is the load-bearing wall beneath essentially all of applied analysis. In optimization theory and machine learning, the Extreme Value Theorem is the reason a loss function continuous on a compact constraint set (e.g. a probability simplex, or a bounded parameter box) is guaranteed to have a global minimum to search for — without compactness, gradient descent could chase a supremum that is never attained. In physics, continuity of fields — temperature T(x,y,z)T(x,y,z), electric potential, fluid velocity — is the baseline assumption needed before one can even ask whether a Partial Derivatives exists at a point, since differentiability at a point requires continuity there as a prerequisite. In numerical analysis, root-finding and fixed-point algorithms in Rn\mathbb{R}^n (e.g. multivariable Newton's method) rely on continuity to guarantee that small input perturbations produce small output perturbations, making iterative schemes stable. In economics, continuity of utility and production functions over compact commodity or feasibility sets is exactly what guarantees (via this same Extreme Value Theorem) that an optimal consumption bundle or profit-maximizing production plan exists at all. In topology and geometric analysis, the componentwise criterion generalizes to continuity of maps into product spaces, a cornerstone fact used constantly when working with vector fields, parametrized curves and surfaces, and coordinate charts on manifolds.

Ask the tutor: Why does agreement of a multivariable limit along every straight line through a point still fail to guarantee the limit exists?

DefinitionLimit and continuity for functions Rn→Rm\mathbb{R}^n \to \mathbb{R}^m

Let D⊆RnD \subseteq \mathbb{R}^n, let aa be a limit point of DD (i.e., every ball around aa contains a point of DD other than aa), and let f:D→Rmf: D \to \mathbb{R}^m. We say

lim⁡x→af(x)=L\lim_{x \to a} f(x) = L

for some L∈RmL \in \mathbb{R}^m if for every ε>0\varepsilon > 0 there exists δ>0\delta > 0 such that

x∈D,0<∥x−a∥<δ  ⟹  ∥f(x)−L∥<ε,x \in D,\quad 0 < \|x - a\| < \delta \implies \|f(x) - L\| < \varepsilon,

where ∥⋅∥\|\cdot\| denotes the Euclidean norm on the appropriate space (Rn\mathbb{R}^n on the left, Rm\mathbb{R}^m on the right). If a∈Da \in D, we say ff is continuous at aa if

lim⁡x→af(x)=f(a),\lim_{x \to a} f(x) = f(a),

meaning: for every ε>0\varepsilon>0 there is δ>0\delta>0 such that x∈Dx \in D and ∥x−a∥<δ\|x-a\|<\delta imply ∥f(x)−f(a)∥<ε\|f(x)-f(a)\|<\varepsilon (note the ball around aa need not be punctured here, since x=ax = a trivially satisfies the conclusion). We say ff is continuous on DD if it is continuous at every point of DD.

Intuition

In one variable, "x→ax \to a" means xx slides in along the real line from the left or from the right — two directions, period. In Rn\mathbb{R}^n, the set of points within distance δ\delta of aa (minus aa itself) is a punctured ball, and xx can approach aa from any direction on the unit sphere, or along a curve that spirals through infinitely many directions as it closes in. The ε\varepsilon–δ\delta definition demands one radius δ\delta that simultaneously controls ff's behavior on the entire punctured ball — not just along one ray. This is why "plug in along a line y=mxy = mx" is such a seductive but dangerous technique. If f(x,mx)f(x,mx) tends to a limit depending on mm, we instantly know the true limit does not exist, because a single δ\delta-ball must give one value LL for all approach directions, and here different lines already disagree. But if f(x,mx)→Lf(x,mx) \to L for every mm (and even every polynomial path), that alone does not prove the two-variable limit is LL: there might be some non-algebraic curve — for instance x=y2x = y^2 — hiding in the ball where ff misbehaves, invisible to any straight-line test. The first figure shows the punctured disk around aa that the definition quantifies over, together with several paths of approach. The second figure is the classic function f(x,y)=xy2x2+y4f(x,y) = \dfrac{xy^2}{x^2+y^4}, which is constant 00 along every line through the origin but tends to 12\tfrac12 along the parabola x=y2x = y^2 — the canonical illustration that "all lines agree" is not enough.

Figure
The punctured δ\delta-ball around a=(1,1)a=(1,1), approached along a line, a parabola, and a spiral
a = (1,1) δ-ball line path parabola path spiral path x y
Figure
f(x,y)=xy2x2+y4f(x,y)=\dfrac{xy^2}{x^2+y^4} along lines y=mxy=mx (value 0) versus along the parabola x=y2x=y^2 (value 1/2)
x y y = x (f → 0) y = -x (f → 0) x = y² (f → 1/2) (0,0)

Central results

TheoremComponentwise criterion for limits and continuity

Let D⊆RnD \subseteq \mathbb{R}^n, let aa be a limit point of DD, and let f:D→Rmf: D \to \mathbb{R}^m be written in coordinates as f(x)=(f1(x),…,fm(x))f(x) = \big(f_1(x), \dots, f_m(x)\big) where each fi:D→Rf_i : D \to \mathbb{R}. Let L=(L1,…,Lm)∈RmL = (L_1, \dots, L_m) \in \mathbb{R}^m. Then

lim⁡x→af(x)=L  ⟺  lim⁡x→afi(x)=Li for every i=1,…,m.\lim_{x \to a} f(x) = L \quad \iff \quad \lim_{x \to a} f_i(x) = L_i \text{ for every } i = 1,\dots,m.

Consequently, if a∈Da \in D, ff is continuous at aa if and only if every component function fif_i is continuous at aa.

Proof

We use the elementary inequality relating the Euclidean norm on Rm\mathbb{R}^m to its coordinates: for v=(v1,…,vm)∈Rmv = (v_1,\dots,v_m) \in \mathbb{R}^m,

∣vi∣≤∥v∥≤∑j=1m∣vj∣for each i=1,…,m.(∗)|v_i| \leq \|v\| \leq \sum_{j=1}^m |v_j| \qquad \text{for each } i = 1,\dots,m. \tag{$\ast$}

The left inequality holds because ∥v∥2=∑jvj2≥vi2\|v\|^2 = \sum_j v_j^2 \geq v_i^2. The right inequality holds because ∥v∥=∑jvj2≤(∑j∣vj∣)2=∑j∣vj∣\|v\| = \sqrt{\sum_j v_j^2} \leq \sqrt{\left(\sum_j |v_j|\right)^2} = \sum_j |v_j|, using that all terms are nonnegative. (⇒\Rightarrow) Suppose lim⁡x→af(x)=L\lim_{x\to a} f(x) = L. Fix i∈{1,…,m}i \in \{1,\dots,m\} and let ε>0\varepsilon > 0. By hypothesis there is δ>0\delta > 0 such that 0<∥x−a∥<δ0 < \|x-a\| < \delta (with x∈Dx \in D) implies ∥f(x)−L∥<ε\|f(x) - L\| < \varepsilon. For such xx, apply the left inequality in (∗)(\ast) to the vector v=f(x)−Lv = f(x) - L, whose ii-th coordinate is fi(x)−Lif_i(x) - L_i:

∣fi(x)−Li∣≤∥f(x)−L∥<ε.|f_i(x) - L_i| \leq \|f(x) - L\| < \varepsilon.

Thus for every ε>0\varepsilon>0 this same δ\delta works, so lim⁡x→afi(x)=Li\lim_{x\to a} f_i(x) = L_i. (⇐\Leftarrow) Suppose lim⁡x→afi(x)=Li\lim_{x\to a} f_i(x) = L_i for every i=1,…,mi = 1,\dots,m. Let ε>0\varepsilon > 0. Set ε′=ε/m>0\varepsilon' = \varepsilon/m > 0. For each ii, by hypothesis there is δi>0\delta_i > 0 such that x∈Dx \in D, 0<∥x−a∥<δi0 < \|x-a\| < \delta_i implies ∣fi(x)−Li∣<ε′|f_i(x) - L_i| < \varepsilon'. Let

δ=min⁡(δ1,…,δm)>0\delta = \min(\delta_1, \dots, \delta_m) > 0

(a finite minimum of positive numbers is positive). If x∈Dx \in D and 0<∥x−a∥<δ0 < \|x - a\| < \delta, then 0<∥x−a∥<δi0 < \|x-a\| < \delta_i for every ii, so ∣fi(x)−Li∣<ε′|f_i(x) - L_i| < \varepsilon' for every ii. Apply the right inequality in (∗)(\ast) to v=f(x)−Lv = f(x) - L:

∥f(x)−L∥≤∑i=1m∣fi(x)−Li∣<∑i=1mε′=m⋅εm=ε.\|f(x) - L\| \leq \sum_{i=1}^m |f_i(x) - L_i| < \sum_{i=1}^m \varepsilon' = m \cdot \frac{\varepsilon}{m} = \varepsilon.

Since ε>0\varepsilon > 0 was arbitrary and we exhibited a working δ\delta, we conclude lim⁡x→af(x)=L\lim_{x\to a} f(x) = L. The continuity statement follows immediately: continuity at a∈Da \in D is exactly the limit statement with L=f(a)L = f(a), i.e. Li=fi(a)L_i = f_i(a) for each ii, so ff continuous at aa   ⟺  \iff lim⁡x→af(x)=f(a)\lim_{x\to a} f(x) = f(a)   ⟺  \iff lim⁡x→afi(x)=fi(a)\lim_{x \to a} f_i(x) = f_i(a) for all ii   ⟺  \iff each fif_i is continuous at aa. ■\blacksquare

∎
LemmaPath/sequential characterization of limits

Let D⊆RnD \subseteq \mathbb{R}^n, aa a limit point of DD, f:D→Rmf : D \to \mathbb{R}^m, L∈RmL \in \mathbb{R}^m. Then lim⁡x→af(x)=L\lim_{x\to a} f(x) = L if and only if for every sequence (xk)(x_k) in D∖{a}D \setminus \{a\} with xk→ax_k \to a, we have f(xk)→Lf(x_k) \to L.

Proof

(⇒\Rightarrow) Suppose lim⁡x→af(x)=L\lim_{x\to a} f(x) = L and let (xk)⊂D∖{a}(x_k) \subset D\setminus\{a\} with xk→ax_k \to a. Let ε>0\varepsilon>0; choose δ>0\delta>0 from the definition of the limit. Since xk→ax_k \to a, there is NN such that k≥N  ⟹  ∥xk−a∥<δk \geq N \implies \|x_k - a\| < \delta; since also xk≠ax_k \neq a we have 0<∥xk−a∥<δ0 < \|x_k - a\| < \delta, so ∥f(xk)−L∥<ε\|f(x_k) - L\| < \varepsilon for all k≥Nk \geq N. Hence f(xk)→Lf(x_k) \to L. (⇐\Leftarrow, contrapositive) Suppose lim⁡x→af(x)≠L\lim_{x\to a} f(x) \neq L, i.e. the ε\varepsilon–δ\delta condition fails: there exists ε0>0\varepsilon_0 > 0 such that no δ>0\delta > 0 works, meaning for every δ>0\delta > 0 there is some x∈Dx \in D with 0<∥x−a∥<δ0 < \|x-a\|<\delta but ∥f(x)−L∥≥ε0\|f(x) - L\| \geq \varepsilon_0. Apply this with δ=1/k\delta = 1/k for each k=1,2,3,…k = 1,2,3,\dots to obtain a point xk∈D∖{a}x_k \in D\setminus\{a\} with 0<∥xk−a∥<1/k0 < \|x_k - a\| < 1/k and ∥f(xk)−L∥≥ε0\|f(x_k) - L\| \geq \varepsilon_0. Then xk→ax_k \to a (since ∥xk−a∥→0\|x_k-a\|\to 0) but f(xk)↛Lf(x_k) \not\to L (since ∥f(xk)−L∥\|f(x_k)-L\| never drops below ε0\varepsilon_0). This is a sequence violating the right-hand side, proving the contrapositive. ■\blacksquare

∎
CorollaryTwo-path test for non-existence

If (xk)→a(x_k) \to a and (yk)→a(y_k) \to a are two sequences in D∖{a}D \setminus\{a\} with f(xk)→L1f(x_k) \to L_1 and f(yk)→L2f(y_k) \to L_2 where L1≠L2L_1 \neq L_2, then lim⁡x→af(x)\lim_{x\to a} f(x) does not exist. This follows immediately from the Path/Sequential Characterization lemma: if the limit LL existed, both sequential limits would have to equal LL, forcing L1=L2L_1 = L_2, a contradiction.

TheoremExtreme Value Theorem in several variables

Let K⊆RnK \subseteq \mathbb{R}^n be nonempty and compact (equivalently, by the Heine–Borel Theorem, closed and bounded), and let f:K→Rf : K \to \mathbb{R} be continuous on KK. Then ff is bounded on KK, and there exist points p,q∈Kp, q \in K such that

f(p)=sup⁡x∈Kf(x),f(q)=inf⁡x∈Kf(x).f(p) = \sup_{x \in K} f(x), \qquad f(q) = \inf_{x \in K} f(x).

That is, ff attains a global maximum and a global minimum value on KK.

Proof

We give the standard sequential-compactness proof.

Step 1: KK is sequentially compact. By the Heine–Borel theorem (established alongside Open and Closed Sets in R^n), a subset of Rn\mathbb{R}^n is compact if and only if it is closed and bounded, and closed-and-bounded subsets of Rn\mathbb{R}^n are sequentially compact: every sequence in KK has a subsequence converging to a point of KK. We use this fact.

Step 2: ff is bounded above on KK. Suppose for contradiction ff is not bounded above. Then for each k∈Nk \in \mathbb{N} there exists xk∈Kx_k \in K with f(xk)>kf(x_k) > k. By sequential compactness, (xk)(x_k) has a subsequence (xkj)(x_{k_j}) converging to some p∈Kp \in K. Since ff is continuous at pp and xkj→px_{k_j} \to p, the sequential characterization of continuity (the special case L=f(a)L = f(a) of the Path/Sequential Characterization lemma above) gives f(xkj)→f(p)f(x_{k_j}) \to f(p), a finite real number. But f(xkj)>kj→∞f(x_{k_j}) > k_j \to \infty, so f(xkj)→∞f(x_{k_j}) \to \infty, contradicting convergence to the finite value f(p)f(p). Hence ff is bounded above on KK. The same argument applied to −f-f shows ff is bounded below.

Step 3: The supremum is attained. Let M=sup⁡x∈Kf(x)M = \sup_{x \in K} f(x), which is a finite real number by Step 2 and the completeness of R\mathbb{R} (a nonempty set bounded above has a supremum). By the definition of supremum, for each k∈Nk \in \mathbb{N} there exists xk∈Kx_k \in K with

M−1k<f(xk)≤M.M - \frac{1}{k} < f(x_k) \leq M.

This produces a sequence (xk)(x_k) in KK with f(xk)→Mf(x_k) \to M (squeeze: M−1/k<f(xk)≤MM - 1/k < f(x_k) \le M and both bounds →M\to M). By sequential compactness of KK, some subsequence (xkj)(x_{k_j}) converges to a point p∈Kp \in K. By continuity of ff at pp,

f(p)=lim⁡j→∞f(xkj).f(p) = \lim_{j\to\infty} f(x_{k_j}).

But (f(xkj))(f(x_{k_j})) is a subsequence of (f(xk))(f(x_k)), which converges to MM; every subsequence of a convergent sequence converges to the same limit, so f(xkj)→Mf(x_{k_j}) \to M as well. By uniqueness of limits in R\mathbb{R}, f(p)=Mf(p) = M. Thus the supremum is attained at p∈Kp \in K.

Step 4: The infimum is attained. Apply Steps 2–3 to −f-f, which is continuous on KK since ff is (a constant multiple of a continuous function is continuous — immediate from the ε\varepsilon–δ\delta definition with ε\varepsilon rescaled). This produces q∈Kq \in K with −f(q)=sup⁡x∈K(−f(x))=−inf⁡x∈Kf(x)-f(q) = \sup_{x\in K}(-f(x)) = -\inf_{x \in K} f(x), i.e. f(q)=inf⁡x∈Kf(x)f(q) = \inf_{x\in K} f(x). ■\blacksquare

∎

Worked examples

Example 1 (a limit that genuinely exists, verified rigorously — not just "by inspection"). Let

f(x,y)=x3+2y3x2+y2,(x,y)≠(0,0).f(x,y) = \frac{x^3 + 2y^3}{x^2 + y^2}, \qquad (x,y) \neq (0,0).

Claim: lim⁡(x,y)→(0,0)f(x,y)=0\displaystyle\lim_{(x,y)\to(0,0)} f(x,y) = 0. Path tests first (necessary but not sufficient — used here to build confidence, not as proof). Along y=0y = 0: f(x,0)=x3/x2=x→0f(x,0) = x^3/x^2 = x \to 0. Along y=xy=x: f(x,x)=3x3/(2x2)=32x→0f(x,x) = 3x^3/(2x^2) = \tfrac32 x \to 0. Along y=mxy = mx: f(x,mx)=x3(1+2m3)x2(1+m2)=1+2m31+m2 x→0f(x,mx) = \dfrac{x^3(1+2m^3)}{x^2(1+m^2)} = \dfrac{1+2m^3}{1+m^2}\,x \to 0 as x→0x\to 0, for every fixed mm. All lines agree on 00. This is consistent with, but does not prove, the claim. Rigorous proof via the ε\varepsilon–δ\delta definition. We bound ∣f(x,y)∣|f(x,y)| using polar-type estimates. Write r=x2+y2=∥(x,y)∥r = \sqrt{x^2+y^2} = \|(x,y)\|. Since ∣x∣≤r|x| \leq r and ∣y∣≤r|y| \leq r,

∣x3+2y3∣≤∣x∣3+2∣y∣3≤r3+2r3=3r3.|x^3 + 2y^3| \leq |x|^3 + 2|y|^3 \leq r^3 + 2r^3 = 3r^3.

Also x2+y2=r2x^2+y^2 = r^2. Hence for (x,y)≠(0,0)(x,y)\neq(0,0),

∣f(x,y)∣=∣x3+2y3∣x2+y2≤3r3r2=3r=3∥(x,y)∥.|f(x,y)| = \frac{|x^3+2y^3|}{x^2+y^2} \leq \frac{3r^3}{r^2} = 3r = 3\|(x,y)\|.

Given ε>0\varepsilon > 0, choose δ=ε/3\delta = \varepsilon/3. If 0<∥(x,y)−(0,0)∥<δ0 < \|(x,y) - (0,0)\| < \delta, then r<ε/3r < \varepsilon/3, so

∣f(x,y)−0∣≤3r<3⋅ε3=ε.|f(x,y) - 0| \leq 3r < 3\cdot\frac{\varepsilon}{3} = \varepsilon.

This holds for every direction of approach simultaneously — the bound 3r3r depends only on the distance rr to the origin, not on the angle — so the ε\varepsilon–δ\delta condition is verified and

lim⁡(x,y)→(0,0)f(x,y)=0.■\lim_{(x,y)\to(0,0)} f(x,y) = 0. \qquad \blacksquare

Example 2 (a limit that does not exist, with a full two-path disproof plus a check that the "all lines agree" trap is avoided). Let

g(x,y)=x2yx4+y2,(x,y)≠(0,0).g(x,y) = \frac{x^2 y}{x^4 + y^2}, \qquad (x,y) \neq (0,0).

Step 1: check lines. Along y=mxy = mx (mm fixed, x≠0x \neq 0):

g(x,mx)=x2(mx)x4+m2x2=mx3x2(x2+m2)=mxx2+m2→x→00g(x, mx) = \frac{x^2(mx)}{x^4 + m^2x^2} = \frac{m x^3}{x^2(x^2+m^2)} = \frac{mx}{x^2+m^2} \xrightarrow[x\to 0]{} 0

for every fixed mm (including m=0m=0, giving identically 00). So every straight-line path gives limit 00. If we stopped here we would (wrongly) conjecture the limit is 00. Step 2: try the curve y=x2y = x^2, matching the denominator's degree structure. For x≠0x \neq 0,

g(x,x2)=x2⋅x2x4+x4=x42x4=12.g(x, x^2) = \frac{x^2 \cdot x^2}{x^4 + x^4} = \frac{x^4}{2x^4} = \frac{1}{2}.

So along the parabola y=x2y = x^2, gg is constant at 12\tfrac12, for all x≠0x \neq 0 — in particular g(x,x2)→12g(x,x^2) \to \tfrac12 as x→0x \to 0. Step 3: apply the Two-Path Test corollary. Take xk=(1/k, 0)x_k = (1/k,\,0), so xk→(0,0)x_k \to (0,0) and g(xk)=0→0=:L1g(x_k) = 0 \to 0 =: L_1. Take yk=(1/k, 1/k2)y_k = (1/k,\,1/k^2), so yk→(0,0)y_k \to (0,0) and g(yk)=12→12=:L2g(y_k) = \tfrac12 \to \tfrac12 =: L_2. Since L1=0≠12=L2L_1 = 0 \neq \tfrac12 = L_2, the corollary to the Path/Sequential Characterization lemma applies directly:

lim⁡(x,y)→(0,0)g(x,y) does not exist.\lim_{(x,y)\to(0,0)} g(x,y) \text{ does not exist.}

This example is the sharpest possible illustration of the danger in Example 1's method: infinitely many straight-line checks (an entire pencil of lines) all agreeing is still not a proof, because the true obstruction lives along a curve tangent to every line at the origin.

CounterexampleDropping "the same δ\delta works for all directions" — why finitely-many-paths agreement isn't the definition

The theorem implicitly used in practice — "if ff tends to LL along every straight line through aa, then lim⁡x→af(x)=L\lim_{x\to a} f(x) = L" — is false, and Example 2 already disproves it. Let us isolate exactly which hypothesis is missing and restate the false claim precisely to see the gap. False claim: If for every m∈Rm \in \mathbb{R}, lim⁡x→0f(x,mx)=L\lim_{x\to 0} f(x, mx) = L (and also lim⁡y→0f(0,y)=L\lim_{y\to 0} f(0,y) = L), then lim⁡(x,y)→(0,0)f(x,y)=L\lim_{(x,y)\to(0,0)} f(x,y) = L. Why this looks plausible: it is tempting to think that since every direction from the origin is some line y=mxy = mx (or the vertical line x=0x=0), checking "all mm" checks "all directions," hence should check "all paths." Where it fails: the definition of lim⁡x→af(x)=L\lim_{x\to a} f(x) = L requires one single δ\delta that controls ff on the full punctured ball 0<∥x−a∥<δ0 < \|x - a\| < \delta, uniformly over the entire ball at once. Checking a line y=mxy = mx only samples the restriction of ff to that one ray; the δ\delta that works for that ray can depend on mm, and as mm (or the shape of the approach curve) varies continuously, no single δ\delta need work for all of them simultaneously — especially for curves, like parabolas, that are not lines at all and hence aren't sampled by the family {y=mx}\{y=mx\}. Concretely, using g(x,y)=x2yx4+y2g(x,y) = \dfrac{x^2y}{x^4+y^2} from Example 2: every line y=mxy=mx gives limit 00 (verified above), satisfying every hypothesis of the false claim with L=0L=0. Yet the true limit does not exist, since g(x,x2)≡12g(x,x^2) \equiv \tfrac12 for x≠0x \neq 0 while g(x,0)≡0g(x,0) \equiv 0. The hypothesis silently dropped is that the family of paths tested — lines through the origin — does not exhaust "all possible modes of approach": it misses curves such as y=x2y = x^2 that hug the origin in a way no line does. The moral: a path test can only be used to disprove a limit (via the rigorously proven Two-Path Test corollary); it can never be used to prove one, because "all lines" is not "all sequences/paths."

Scope, edge cases & common traps

SituationWhat to checkCommon mistake
Suspect limit existsBound ∣f(x)−L∣\lvert f(x)-L\rvert by a function of r=∥x−a∥r=\|x-a\| alone (polar-type or algebraic estimate), then do ε\varepsilon–δ\deltaTreating agreement along several lines as a proof
Suspect limit fails to existFind two paths (sequences) to aa giving different limiting valuesOnly trying lines y=mxy=mx, missing curves like y=x2y=x^2 or x=y3x = y^3
Vector-valued f:Rn→Rmf:\mathbb R^n\to\mathbb R^mApply componentwise criterion to reduce to mm scalar limitsForgetting the criterion requires the same point aa and same limit point status for all components — it does not let you take limits along different paths per component
Extreme Value TheoremConfirm domain KK is both closed and bounded (compact), and ff continuous on all of KKApplying EVT on an open or unbounded set (e.g. open disk, or f(x,y)=xf(x,y)=x on all of R2\mathbb R^2), where no maximum exists
Continuity at a boundary/limit point not in domainThe definition of continuity requires a∈Da \in DConfusing "limit exists at aa" with "ff is continuous at aa" when ff isn't even defined there, or when f(a)f(a) is defined but disagrees with the limit (removable discontinuity)

The mathematician behind it

Origins · 1 minKarl Weierstrass and how this was discovered
KW
Karl Weierstrass
1815–1897 · Prussia/German Empire, foundations of analysis

Karl Weierstrass is the figure most responsible for the ε\varepsilon–δ\delta formalism that makes rigorous multivariable limits possible at all. Working as a Gymnasium teacher for over a decade before his mathematical talent was recognized, Weierstrass later became a professor in Berlin, where his lectures — never published by him directly, but transcribed and circulated by students — set the standard for rigor that all of modern real and complex analysis still follows. He insisted that "limit," "continuity," and "convergence" be defined by explicit, checkable inequalities rather than appeals to infinitesimals or geometric intuition, banishing centuries of ambiguity about what a limit "really is." His construction of a continuous, nowhere-differentiable function (built from a lacunary trigonometric series) shocked contemporaries who believed continuity implied differentiability except at isolated points, and it is a direct descendant of the same ε\varepsilon–δ\delta machinery used here — a striking early demonstration that continuity alone, without further hypotheses, is a far weaker property than intuition suggests.

History
- 1817 — Bernard Bolzano gives an early rigorous definition of continuity and proves the intermediate value theorem for one variable, largely unnoticed at the time.
- 1821 — Augustin-Louis Cauchy's Cours d'analyse popularizes limit-based definitions of continuity and convergence, though still informally phrased.
- 1861 — Weierstrass's Berlin lectures introduce the fully symbolic ε\varepsilon–δ\delta definition of limit and continuity that becomes the permanent standard.
- 1872 — Weierstrass exhibits a continuous nowhere-differentiable function, showing continuity does not imply local "niceness," a cautionary tale extended by pathological multivariable limits like x2y/(x4+y2)x^2y/(x^4+y^2).
- 1906 — Maurice Fréchet develops general metric space theory, giving the modern language (ε\varepsilon–δ\delta in normed spaces Rn\mathbb{R}^n) in which multivariable limits are now routinely stated.
Why isn't checking limits along the x-axis and y-axis enough, like it was implicitly enough in one variable?

In one variable, a point on R\mathbb{R} truly has only two sides, so checking the left-hand and right-hand limits genuinely exhausts all ways of approaching a point. In Rn\mathbb{R}^n for n≥2n \geq 2, the "boundary" of a small ball around a point is an entire sphere (a circle in R2\mathbb{R}^2, an actual sphere in R3\mathbb{R}^3), containing infinitely many directions, plus infinitely many curved paths that are not straight lines at all. Checking the two axes samples only two of infinitely many rays, and rays alone don't sample curved paths like parabolas or spirals. The only way to prove a multivariable limit exists is to directly verify the ε\varepsilon–δ\delta definition (often via an inequality bounding ∣f(x)−L∣|f(x)-L| by a function of ∥x−a∥\|x-a\| alone), since that definition quantifies over the entire punctured ball uniformly.

Is it true that continuity in each variable separately implies continuity as a function of both variables together?

No, and this is one of the most important cautionary facts in the subject. Define h(x,y)=xyx2+y2h(x,y) = \dfrac{xy}{x^2+y^2} for (x,y)≠(0,0)(x,y)\neq(0,0) and h(0,0)=0h(0,0)=0. Fixing y=0y=0, h(x,0)=0h(x,0)=0 for all xx, which is continuous in xx; fixing x=0x=0, h(0,y)=0h(0,y)=0 for all yy, continuous in yy. So hh is continuous in each variable separately at the origin. But along the line y=xy=x, h(x,x)=x2/(2x2)=1/2h(x,x) = x^2/(2x^2) = 1/2 for all x≠0x\neq0, which does not tend to h(0,0)=0h(0,0)=0. So hh is not (jointly) continuous at (0,0)(0,0). Joint continuity is a strictly stronger condition than separate continuity in each variable.

Does the componentwise criterion mean I can compute limits of vector-valued functions one coordinate at a time using ordinary single-variable rules?

Partially — the criterion legitimately reduces the vector limit problem to mm scalar limit problems, but each of those scalar problems is still a genuine multivariable limit (since the domain point aa is still in Rn\mathbb{R}^n), not a single-variable limit unless n=1n=1. For example, for f(x,y)=(x2+y, sin⁡(xy)/(x2+1))f(x,y) = (x^2+y, \, \sin(xy)/(x^2+1)) mapping R2→R2\mathbb{R}^2 \to \mathbb{R}^2, the criterion says lim⁡(x,y)→(a,b)f(x,y)\lim_{(x,y)\to(a,b)} f(x,y) exists iff both lim⁡(x,y)→(a,b)(x2+y)\lim_{(x,y)\to(a,b)} (x^2+y) and lim⁡(x,y)→(a,b)sin⁡(xy)/(x2+1)\lim_{(x,y)\to(a,b)} \sin(xy)/(x^2+1) exist — and each of those must still be handled with genuinely multivariable techniques (continuity of elementary functions, algebraic limit laws, or ε\varepsilon–δ\delta estimates), just one coordinate at a time instead of all at once.

Why does the Extreme Value Theorem fail on an open set, like the open unit disk?

Consider f(x,y)=11−x2−y2f(x,y) = \dfrac{1}{1-x^2-y^2} on the open unit disk D={(x,y):x2+y2<1}D = \{(x,y): x^2+y^2<1\}, which is bounded but not closed (it's missing its boundary circle). This ff is continuous everywhere on DD, but as (x,y)(x,y) approaches the boundary circle from inside, f(x,y)→∞f(x,y) \to \infty; since the boundary points are excluded from the domain, ff is unbounded on DD and certainly attains no maximum. The proof of the Extreme Value Theorem crucially used sequential compactness to guarantee that a maximizing sequence has a subsequence converging to a point of KK itself; on a non-closed set, that limit point can escape the domain, which is exactly the failure mode here.

If a function is continuous at every point of an open set, is it automatically bounded there?

No — continuity is a purely local, pointwise property and carries no global information by itself. The function f(x,y)=1/(x2+y2)f(x,y) = 1/(x^2+y^2) is continuous at every point of the punctured plane R2∖{(0,0)}\mathbb{R}^2\setminus\{(0,0)\} (an open set), yet is unbounded near the origin. The Extreme Value Theorem's boundedness conclusion crucially requires the domain to be compact (closed and bounded); it is precisely the compactness hypothesis, not continuity alone, that forces a continuous function to be globally bounded and to attain its extrema.

What is the relationship between the ε\varepsilon–δ\delta definition and the "open sets" definition of continuity I may have seen?

For f:Rn→Rmf: \mathbb{R}^n \to \mathbb{R}^m, the ε\varepsilon–δ\delta definition given here is provably equivalent to the topological definition: ff is continuous on Rn\mathbb{R}^n iff f−1(U)f^{-1}(U) is open in Rn\mathbb{R}^n for every open U⊆RmU \subseteq \mathbb{R}^m. The equivalence is proved by unwinding definitions: given ε\varepsilon–δ\delta continuity, for open UU and a∈f−1(U)a \in f^{-1}(U), pick a ball B(f(a),ε)⊆UB(f(a),\varepsilon)\subseteq U (possible since UU is open), get δ\delta from continuity, and check B(a,δ)⊆f−1(U)B(a,\delta)\subseteq f^{-1}(U); the converse runs the same argument backward using that balls are open sets. This equivalence, built on Open and Closed Sets in R^n, is why "continuous" can be characterized either analytically (via inequalities) or topologically (via preimages of opens), and both formulations are used throughout analysis and topology.

Before this

Where this leads