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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:

  1. Direct substitution whenever the expression is built from continuous pieces (polynomials, sin⁡\sin, cos⁡\cos, e(⋅)e^{(\cdot)}, log⁡\log, ...) combined algebraically or by composition, with any denominator nonzero at the target point. …
Figure 8.7Fig. 8.7 — interplay of $\varepsilon$ and $\delta$ for $f(x)=\frac{x^{2}-9}{x-3},\ x\neq 3$ (the line $y=x+3$ with a hole at $(3,6)$); the limit as $x\to 3$ is $6$
Fig. 8.7 — Fig. 8.7 — interplay of $\varepsilon$ and $\delta$ for $f(x)=\frac{x^{2}-9}{x-3},\ x\neq 3$ (the line $y=x+3$ with a hole at $(3,6)$); the limit as $x\to 3$ is $6$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig. 8.7 — interplay of ε\varepsilon and δ\delta for f(x)=x2−9x−3, x≠3f(x)=\frac{x^{2}-9}{x-3},\ x\neq 3 (the line y=x+3y=x+3 with a hole at (3,6)(3,6)); the limit as …

Figure 8.8Fig. 8.8 — interplay of $\varepsilon$ and $\delta$ for $f(x)=\frac{x^{3}-8}{x-2},\ x\neq 2$ (the parabola $y=x^{2}+2x+4$ with a hole at $(2,12)$); the limit as $x\to 2$ is $12$
Fig. 8.8 — Fig. 8.8 — interplay of $\varepsilon$ and $\delta$ for $f(x)=\frac{x^{3}-8}{x-2},\ x\neq 2$ (the parabola $y=x^{2}+2x+4$ with a hole at $(2,12)$); the limit as $x\to 2$ is $12$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig. 8.8 — interplay of ε\varepsilon and δ\delta for f(x)=x3−8x−2, x≠2f(x)=\frac{x^{3}-8}{x-2},\ x\neq 2 (the parabola y=x2+2x+4y=x^{2}+2x+4 with a hole at (2,12)(2,12)); the limit as …

Figure 8.9Fig 8.9 Limit of a function: a neighbourhood of (u,v) in the domain R^2 maps under F into the interval (L-e, L+e) on the real line, illustrating the limit L
Fig. 8.9 — Fig 8.9 Limit of a function: a neighbourhood of (u,v) in the domain R^2 maps under F into the interval (L-e, L+e) on the real line, illustrating the limit L

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig 8.9 Limit of a function: a neighbourhood of (u,v) in the domain R^2 maps under F into the interval (L-e, L+e) on the real line, illustrati …

Figure 8.10Fig 8.10 Continuity of a function: a neighbourhood of (u,v) maps under F into (f(u,v)-e, f(u,v)+e), with the limit value equal to f(u,v)
Fig. 8.10 — Fig 8.10 Continuity of a function: a neighbourhood of (u,v) maps under F into (f(u,v)-e, f(u,v)+e), with the limit value equal to f(u,v)

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig 8.10 Continuity of a function: a neighbourhood of (u,v) maps under F into (f(u,v)-e, f(u,v)+e), with the limit value equa …