Concept understanding — Functions of Several Variables & Limits
A function of one variable y=f(x) traces a curve in the xy-plane. A function of two variables F(x,y) is visualized by graphing z=F(x,y), which traces a surface in xyz-space: the point (x,y,F(x,y)) sits F(x,y) units above (or below) the point (x,y) in the xy-plane.
Fixing one variable slices the surface with a plane and produces a curve: for g(x,y)=30−x2−y2, holding y=3 gives g(x,3)=21−x2 (a parabola, the intersection of the surface with the plane y=3), and holding x=2 gives g(2,y)=26−y2. The surface z=30−x2−y2 itself is called a paraboloid. This is the natural generalization from one to several variables: profit as a function of the units of two products, or volume as a function of length, width and height, are genuinely functions of more than one variable and cannot be reduced to a single-variable picture.
Neighbourhoods in R2. To define limits and continuity for F(x,y), replace the one-variable interval-neighbourhood (x0−δ,x0+δ) by an open disc:
Br((u,v))={(x,y)∈R2∣(x−u)2+(y−v)2<r2}
— the set of points strictly within distance r of (u,v). Removing the centre gives a deleted neighbourhood.
Definition (Limit of a Function of Two Variables).F has limit L at (u,v), written (x,y)→(u,v)limF(x,y)=L, if for every neighbourhood (L−ε,L+ε), ε>0, of L there exists a δ-neighbourhood Bδ((u,v)) of (u,v) such that (x,y)∈Bδ((u,v))∖{(u,v)}⇒F(x,y)∈(L−ε,L+ε).
Definition (Continuity).F is continuous at (u,v) if (1) F(u,v) is defined, (2) (x,y)→(u,v)limF(x,y) exists, and (3) that limit equals F(u,v) — exactly the same three-part test as one variable, now over R2. All the standard limit theorems (limits of sums, products, quotients, composition with a continuous function) carry over unchanged from one variable to several.
Watch out
The crucial new subtlety: (x,y) must approach (u,v) along every possible path, not just straight lines, for the limit to exist. A classic failure is f(x,y)=x2+y2xy at the origin — along the line y=mx the value is the constant 1+m2m, which is different for different slopes m, so the two-variable limit does not exist even though every straight-line limit does. Checking finitely many paths can only ever disprove a limit (by finding two paths that disagree); it can never by itself prove the limit exists, since some other, non-linear path (e.g. a parabola y=kx2) might still disagree.
Working method to evaluate a limit / test continuity at (u,v):
If F is built from continuous pieces (polynomials, sin,cos,e(⋅),log, ...) by algebraic combination or composition, and the denominator (if any) is nonzero at (u,v), substitute directly — the limit is F(u,v).
If the expression is a genuine 00 form, try to factor/simplify algebraically (e.g. rationalizing x−y via (x−y)(x+y)=x−y) so the singular factor cancels.
If simplification is not obvious, use a squeeze/bound argument: show ∣F(x,y)−L∣≤ (something that visibly →0), often using 2∣xy∣≤x2+y2 (which follows from (x−y)2≥0).
To show a limit does not exist, compute the limit along two different paths (e.g. y=mx for varying m, or a line vs. a parabola y=kx2) and exhibit disagreement.
Factor: y2−xy=−y(x−y) and x−y=x+yx−y, so f(x,y)=−y(x+y) for x=y.
✓Final answer
f(x,y)=−yx−y3/2→0 as (x,y)→(0,0), so lim=0 — proved.
Rationalize the denominator x−y so the singular factor (x−y) cancels between numerator and denominator, reducing f to a manifestly continuous expression.
Step 1. Factor the numerator.y2−xy=y(y−x)=−y(x−y).
Step 2. Rationalize the denominator. Multiplying x−y by its conjugate gives (x−y)(x+y)=x−y, so
(This cancellation is valid precisely where the original f is defined, i.e. x=y so the original denominator is nonzero.)
Step 4. Take the limit of the simplified expression.f(x,y)=−yx−yy=−yx−y3/2, both terms being continuous functions of (x,y) that vanish as (x,y)→(0,0) (since y→0 and x,y3/2 stay bounded near the origin).