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Mathematics · Ch 8 — Differentials and Partial Derivatives

Limit and Continuity of Functions of Two Variables

8.4

Limit and Continuity of Functions of Two Variables

Definition 8.6 (Limit of a Function of Two Variables). Let A={(x,y)∣a<x<b, c<y<d}⊂R2A=\{(x,y)\mid a<x<b,\,c<y<d\}\subset\mathbb R^2 and F:A→RF:A\to\mathbb R. FF has a limit LL at (u,v)(u,v) if: for every neighbourhood (L−ε,L+ε)(L-\varepsilon,L+\varepsilon), ε>0\varepsilon>0, of LL, there exists a δ\delta-neighbourhood Bδ((u,v))⊂AB_\delta((u,v))\subset A of (u,v)(u,v) such that

(x,y)∈Bδ((u,v))∖{(u,v)} ⟹ F(x,y)∈(L−ε,L+ε).(x,y)\in B_\delta((u,v))\setminus\{(u,v)\} \ \Longrightarrow\ F(x,y)\in(L-\varepsilon,L+\varepsilon).

We write lim⁡(x,y)→(u,v)F(x,y)=L\displaystyle\lim_{(x,y)\to(u,v)}F(x,y)=L if such a limit exists. All the standard limit theorems (limits of sums, differences, products, quotients — where the denominator's limit is nonzero — and composition with a continuous function) that hold for one-variable limits hold, unchanged in form, for functions of several variables.

Definition 8.7 (Continuity). FF is continuous at (u,v)(u,v) if: (1) FF is defined at (u,v)(u,v); (2) lim⁡(x,y)→(u,v)F(x,y)\displaystyle\lim_{(x,y)\to(u,v)}F(x,y) exists; and (3) that limit equals F(u,v)F(u,v) — the same three-part test as one variable, carried over verbatim.

Watch out

The genuinely new subtlety compared to one variable: the values F(x,y)F(x,y) must approach the same LL as (x,y)(x,y) approaches (u,v)(u,v) along every possible path to (u,v)(u,v) — not only along straight lines, but along any curve whatsoever. This is what makes two-variable limits strictly harder to establish than one-variable limits (though, when a limit fails to exist, exhibiting just two disagreeing paths is enough to disprove it).

Worked example — a limit that fails to exist. Consider f(x,y)=xyx2+y2f(x,y)=\dfrac{xy}{x^2+y^2} for (x,y)≠(0,0)(x,y)\ne(0,0), f(0,0)=0f(0,0)=0. Along any straight line y=mxy=mx through the origin,

lim⁡x→0f(x,mx)=lim⁡x→0x(mx)x2+(mx)2=lim⁡x→0mx2x2(1+m2)=m1+m2,\lim_{x\to0}f(x,mx) = \lim_{x\to0}\frac{x(mx)}{x^2+(mx)^2} = \lim_{x\to0}\frac{mx^2}{x^2(1+m^2)} = \frac{m}{1+m^2},

a value that genuinely depends on the slope mm of the approach line — different lines through the origin give different limiting values (e.g. m=0m=0 gives 00, m=1m=1 gives 12\tfrac12). Since the limit is not the same along every path, lim⁡(x,y)→(0,0)f(x,y)\displaystyle\lim_{(x,y)\to(0,0)}f(x,y) does not exist, and ff is consequently not continuous at (0,0)(0,0).

Worked example — establishing continuity via a bound. Let g(x,y)=x2yx2+y2g(x,y)=\dfrac{x^2y}{x^2+y^2} for (x,y)≠(0,0)(x,y)\ne(0,0), g(0,0)=0g(0,0)=0; this is continuous everywhere, including at the origin. Away from the origin it is a quotient of continuous functions with nonvanishing denominator, hence continuous there directly. At (0,0)(0,0):

∣g(x,y)−g(0,0)∣=∣x2yx2+y2∣=x2x2+y2 ∣y∣≤∣y∣,|g(x,y)-g(0,0)| = \left|\frac{x^2y}{x^2+y^2}\right| = \frac{x^2}{x^2+y^2}\,|y| \le |y|,

using x2x2+y2≤1\dfrac{x^2}{x^2+y^2}\le1. Since (x,y)→(0,0)(x,y)\to(0,0) forces ∣y∣→0|y|\to0, the squeeze gives lim⁡(x,y)→(0,0)g(x,y)=0=g(0,0)\displaystyle\lim_{(x,y)\to(0,0)}g(x,y)=0=g(0,0), so gg is continuous at (0,0)(0,0) too — hence continuous on all of R2\mathbb R^2.

Working method summary for a two-variable limit/continuity problem: …