Surfaces, Level Sets and Multivariable Limits

Inputs occupy a region

A scalar function $f(x,y)$ assigns one output to each allowed point of a planar domain. Its graph lies in three-dimensional space. A level curve $f(x,y)=c$ stays in the input plane and joins points with equal output. Domain restrictions remain essential: $\sqrt{9-x^2-y^2}$ exists on the closed disk of radius three, while $\ln(9-x^2-y^2)$ exists only on its interior.

A limit $f(x,y)\to L$ at $(a,b)$ requires outputs to approach $L$ along every allowed approach to that point. Two paths with different limits disprove the limit. Agreement along several selected paths never proves it by itself.

Visual guide

VISUAL GUIDETwo paths give different limiting values

Paths in the input plane

Path y = 0
Path y = x

Values along each path

f(t, 0) = 0
f(t, t) = 1/2

For f(x, y) = xy/(x² + y²), approaching the origin along y = 0 gives 0, while approaching along y = x gives 1/2. The right panel plots the value against path parameter t; differing limits prove the 2D limit does not exist.

Worked example: paths disagree

For $f(x,y)=xy/(x^2+y^2)$ away from the origin, the path $y=0$ gives zero. The path $y=x$ gives $1/2$ for nonzero $x$ . Therefore the origin has no two-variable limit, even though both coordinate axes separately give zero.

Even all straight-line tests can miss a curved-path failure. For $g=x^2y/(x^4+y^2)$ , each line through the origin gives limit zero, but the parabola $y=x^2$ gives $1/2$ . The denominator is positive away from the origin, so both comparisons are legitimate.

Worked example: control every path

For $h=x^2y/(x^2+y^2)$ away from the origin,

|h(x,y)|\le |y|\le\sqrt{x^2+y^2}.

The upper bound tends to zero with the distance to the origin, independent of direction. Squeezing proves the limit is zero. Defining $h(0,0)=0$ makes the function continuous there.

Polar coordinates can organize such bounds, but writing a formula in $r,\theta$ is not sufficient unless the estimate is uniform in angle. A denominator containing a vanishing angular factor may prevent the proposed radial bound.

Practice

Describe the level curves of $x^2+y^2$ for $c>0$ .
Find the domain of $1/(x^2+y^2-1)$ .
Prove $x^2y^2/(x^2+y^2)\to0$ at the origin.

Show worked solutions

Circles centered at the origin with radius $\sqrt c$ .
The plane excluding the unit circle.
Since $4x^2y^2\le(x^2+y^2)^2$ , the magnitude is at most $(x^2+y^2)/4$ , which tends to zero along every path.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.