Functions of several variables

One input can contain several independently specified quantities

A temperature field may depend on two position coordinates; a response may depend on several experimental controls. A function of several variables is still a rule assigning exactly one output to each allowed input. Its input is now an ordered tuple, not a collection of interchangeable numbers.

Write f:D⊆Rⁿ→R for a scalar-valued function, and F:D⊆Rⁿ→Rᵐ for a vector-valued map. The domain D specifies which tuples are allowed; the codomain specifies the kind of output. The image, or range, is the subset actually attained. Distinguish a vector-valued output from several different values assigned to one input.

The earlier vector-space and norm lessons provide language for inputs, outputs and distance. This lesson adds domains, graphs, level sets, limits and continuity before lesson 131 introduces partial derivatives, gradients and Jacobians. The exact examples below use dimensionless coordinates; physical applications must supply units and any scaling used to compare input changes.

Keep the domain attached to the formula

For a real-valued formula, a natural domain includes all real tuples for which its operations are defined. A denominator cannot vanish; a real square root requires a nonnegative radicand; a real logarithm requires a strictly positive argument. A problem may specify a smaller domain for physical or experimental reasons, so the formula alone does not always determine the intended function.

For h(x,y)=√(9−x²−y²), the natural domain is the closed disk x²+y²≤9. Its range is [0,3]: the centre gives 3, the boundary gives 0, and every intermediate height occurs. For ℓ(x,y)=ln(y−x²), the domain is y>x², excluding the parabola itself, and the range is all R because y−x² can be any positive real number.

The rational functions xy/(x²+y²), x²y/(x⁴+y²) and x²y/(x²+y²) are each undefined at (0,0) and defined at every other real pair. Their limiting behaviour at the missing point will differ. Naming the domain hole does not predict whether it can be filled continuously.

Two formulas related by cancellation agree only on the domain where the cancellation is valid. For instance (x²−y²)/(x−y)=x+y when x≠y; the original quotient is undefined on x=y. Defining x+y there produces an extension of the original function, not retroactive values of the quotient.

A graph records inputs as well as outputs

The graph of scalar f:D⊆R²→R is the set of triples (x,y,f(x,y)) in R³. Fixing an allowed pair determines one height. A relation such as z²=9−x²−y² generally gives two heights and is not the graph of a single real function of x,y unless a branch is chosen. The upper hemisphere h uses the nonnegative square-root branch.

For a scalar function on R³, a graph belongs to R⁴. A two- or three-dimensional picture can show slices, projections or level sets, but should state which information is retained. A perspective surface drawing is not a literal display of all coordinates in a higher-dimensional graph.

For F(x,y)=(x+y,x−y,xy), the image is a subset of R³, whereas the graph consists of (x,y,F₁,F₂,F₃) in R⁵. If (a,b,c)=F(x,y), then x=(a+b)/2, y=(a−b)/2 and c=(a²−b²)/4. Thus the image is the surface c=(a²−b²)/4; it is not the five-dimensional graph.

A codomain can contain unattained points. The target (0,0,1) belongs to the stated codomain R³ but not to this image: the first two output coordinates force x=y=0, making the third coordinate 0. This distinction matters when solving an inverse or fitting problem.

A level set lives in the input space

For a scalar f and a real c, its level set is {x∈D:f(x)=c}. In two input dimensions it may be a curve, several curves, a point, a region or the empty set. For a three-variable scalar function it can be a level surface. The equality does not guarantee that a smooth manifold exists at every point.

For the plane p(x,y)=1+2x−y, the c-level is the line y=1+2x−c. For the bowl b(x,y)=x²+2y², positive levels are ellipses x²/c+y²/(c/2)=1; the zero level is the origin, and negative levels are empty. For the saddle s(x,y)=x²−y², the zero level is the union y=x and y=−x; positive and negative levels are hyperbolas opening in different coordinate directions.

For h=√(9−x²−y²), a level c is possible only for 0≤c≤3. Within that interval x²+y²=9−c², giving a circle except that c=3 gives one point. A negative c is impossible even though squaring the equation could produce a circle. Domain and sign restrictions survive algebraic manipulation.

For ℓ=ln(y−x²), the c-level is y=x²+eᶜ. It lies strictly above the excluded boundary parabola. For c=0 the offset is 1, and for c=ln 4 it is 4. The domain boundary y=x² is not another finite logarithm level.

A contour map draws selected level sets in the input plane and labels their values. Equal contour increments describe equal output changes, not equal horizontal distances. Closely spaced levels can suggest rapid change in a direction, but slope and gradient require the definitions in the next lesson. Contour intersections, missing levels and degeneracies need mathematical interpretation rather than a smooth-looking drawing.

A slice is useful evidence with a limited scope

Fixing x=a in a two-variable scalar function gives a one-variable rule y↦f(a,y) on its allowed slice. Its graph is a vertical trace of the surface. A horizontal slice z=c instead gives a level set. A trace and a contour answer different questions even when both are drawn as curves.

For b=x²+2y², the trace y=0 is z=x², while the trace x=0 is z=2y². Its horizontal level c=2 is the ellipse x²+2y²=2. These describe the same function through different restrictions; neither pair of traces alone specifies every possible surface.

More generally an approach curve γ(t) in the domain gives the composite f(γ(t)). The curve must actually approach the target input and stay in D\{a}, excluding the target a itself, for all sufficiently close nonzero parameters. Its limit is a necessary consequence of a joint limit, but success along one curve or a finite collection of curves does not establish the joint result.

When physical inputs have different units, a norm on raw coordinates can mix unlike quantities. Choose meaningful scales or a justified metric before treating a small numerical input distance as a small physical perturbation. In the dimensionless examples here, ρ=√(x²+y²) is the ordinary Euclidean distance from the origin.

Approach through the actual domain

A Euclidean neighbourhood of a∈Rⁿ can be described by ‖x−a‖₂<δ for δ>0. The domain may include the whole neighbourhood, only part of it, or none of it. An interior point has some whole ball inside D; a boundary point has arbitrarily nearby points both in D and outside D.

A limit at a is taken through allowed domain points. We state finite limits at accumulation points of D, where every ball contains a point of D other than a. This condition ensures that the punctured approach is meaningful and the limit, if it exists, is unique. The point a itself need not belong to D.

The point (3,0) is on the boundary of the hemisphere’s disk. The function is defined there, and a relative-domain limit uses only points inside or on that disk. Requiring values at exterior points where the radicand is negative would change the question.

For the logarithm domain y>x², points on y=x² are excluded boundary points. They can still be approached from the domain, but continuity at a point is a property stated at a point of the function’s domain. At an isolated domain point, continuity is automatic in the relative-domain definition, even though there is no punctured-domain limit to specify under our accumulation-point convention.

One δ must work for every sufficiently close allowed input

For f:D⊆Rⁿ→R and an accumulation point a of D, write limₓ→ₐ f(x)=L when for every ε>0 there exists δ>0 such that x∈D and 0<‖x−a‖₂<δ imply |f(x)−L|<ε. For a vector-valued map replace the final absolute value by an output norm. The quantifiers require one δ for all allowed input directions, not a separate favourable δ chosen after an input is inspected.

The strict inequality 0<‖x−a‖ excludes the target input itself. Therefore the limit does not depend on a value assigned at a, if there is one. A formula can have a finite limit at a missing point, and a separately assigned value can agree with or disagree with that limit.

In these finite-dimensional Euclidean spaces, an equivalent test uses sequences: every sequence xₖ∈D with xₖ≠a and xₖ→a must satisfy f(xₖ)→L. If the ε–δ condition fails, some fixed ε permits points arbitrarily close to a whose output error is at least ε; selecting one in each radius 1/k produces a contrary sequence.

A joint limit implies the same result along any valid approach curve or sequence. Two approaches with different output limits disprove existence. Testing many paths or a fine numerical grid can find a counterexample, but cannot by itself satisfy the universal condition. Numerical rounding and a finite plotting window add further limits to such evidence.

Use algebraic limit laws where their hypotheses hold

Finite limits respect sums, scalar multiples and products. A quotient respects the corresponding law if the denominator’s limit is nonzero. Composition with a function continuous at the inner limit also preserves the limit, provided the expressions stay in their stated domains.

Polynomials in several variables are continuous everywhere. A rational function is continuous wherever its denominator is nonzero. The square-root function is continuous on [0,∞) relative to that domain, and the logarithm is continuous on (0,∞). Their multivariable compositions are continuous where the inner expressions meet those domain conditions.

For example the plane 1+2x−y tends to 1+2a−b at (a,b). A direct quantitative bound is |2(x−a)−(y−b)|≤√5√((x−a)²+(y−b)²), by Cauchy–Schwarz. Choosing δ=ε/√5 proves the limit in every direction, rather than relying on two coordinate slices.

At a zero denominator, the quotient law gives no conclusion. Another argument may establish a removable limit, or show path dependence, or show unbounded behaviour. A failed hypothesis is a reason to change the argument, not a proof of one particular outcome.

Different path values rule out a joint limit

Let f(x,y)=xy/(x²+y²) on the punctured plane. Along y=0 and x≠0 the value is 0. Along y=x and x≠0 the value is 1/2. Both approaches tend to (0,0), so no finite joint limit exists there. No choice of f(0,0) can repair that failure.

Both iterated limits nevertheless equal 0. For fixed y≠0, taking x→0 gives 0; for y=0 the permitted x-axis values are also 0. Taking the remaining y-limit gives 0, and the reversed order works the same way. Iterated limits impose sequential restrictions on coordinates; they are not automatically the joint limit.

Checking only the two axes would miss the diagonal behaviour. An image whose colours appear to settle at a few sampled locations can miss it as well. A counterexample needs valid points and a genuine limit along them; a universal positive claim needs control of every sufficiently close allowed input.

Every straight line can agree while a curved path differs

Consider g(x,y)=x²y/(x⁴+y²), again undefined only at the origin. On a fixed nonvertical line y=mx with m≠0, cancellation at x≠0 gives g(x,mx)=mx/(x²+m²), which tends to 0. The line y=0 also gives 0, and the vertical line x=0 gives 0.

Thus every fixed straight line through the origin has limit 0. But on y=kx², x≠0, the value is k/(1+k²). Taking k=1 gives 1/2 and k=−1 gives −1/2. Hence the joint limit does not exist. The order “for each fixed line, take a limit” still does not provide one neighbourhood estimate valid for all points.

Both iterated limits are also 0: hold one coordinate fixed, take the other to zero, then take the remaining limit. This example separates three questions—coordinate-wise iteration, all straight-line approaches and the actual joint approach. None can be interchanged merely because their formulas look similar.

The denominator x⁴+y² treats the x and y scales differently near zero. The parabola y=x² keeps its two terms comparable, making a feature that straight lines can miss. Looking for competing scales is a useful way to design an approach test, but the resulting algebra must still be checked.

A bound can control all approach directions at once

Now let r(x,y)=x²y/(x²+y²) on the punctured plane. Since 0≤x²/(x²+y²)≤1, we have |r(x,y)|≤|y|≤√(x²+y²)=ρ. For any ε>0 choose δ=ε. Every permitted point with 0<ρ<δ then satisfies |r(x,y)|<ε, proving the joint limit is 0.

This argument is stronger than testing paths because the same bound holds on the entire punctured disk. The numerator’s degree suggests decay, but the displayed inequality supplies the proof. A degree comparison without control of a possibly small denominator would not be sufficient in general.

At a boundary point the same principle uses the actual domain. For h=√(9−x²−y²) and a=(3,0), set p=(x,y) in the disk. Then h(p)²=9−‖p‖₂²=(3−‖p‖₂)(3+‖p‖₂)≤6‖p−a‖₂, using ‖p‖₂≤3 and the reverse triangle inequality.

Choosing δ=ε²/6 gives h(p)<ε for all permitted p with 0<‖p−a‖₂<δ. Therefore h tends to 0 at that boundary point relative to the disk. This proves more than examining one inward radial line; it covers all nearby allowed points.

Filling a hole requires the right limit and the right value

A function f:D→R is continuous at a∈D when for every ε>0 some δ>0 makes x∈D and ‖x−a‖₂<δ imply |f(x)−f(a)|<ε. At an accumulation point this is equivalent to the finite limit existing and equalling f(a). A function is continuous on its domain if it is continuous at every domain point, including permitted boundary points in the relative sense.

The original r=x²y/(x²+y²) has no value at the origin, so it is not a function defined there. Extending it by r₀(0,0)=0 gives a continuous function on all R² by the preceding bound. Extending the same punctured formula instead by r₁(0,0)=1 gives a function discontinuous at the origin. The limit of the punctured values remains 0 in either extension.

For f=xy/(x²+y²) or g=x²y/(x⁴+y²), no assigned origin value yields continuity, because the joint limit itself fails. All three original rational functions remain continuous at every nonzero input: the denominators are nonzero there. A local failure at one point is not a failure everywhere.

The hemisphere is continuous throughout its closed disk, including the boundary. The logarithm is continuous throughout y>x² but has no finite continuous extension at (0,0): the valid path (0,t), t>0, gives ln t→−∞. Divergence along this path already excludes a finite limiting value; it is not a domain value at t=0.

Continuity concerns small changes of the whole input, not just small changes along selected slices. It does not by itself establish differentiability, a best linear approximation or a scientifically adequate model. Measurement noise, sharp transitions and a restricted model domain require their own interpretation.

Several output components share one limiting requirement

A vector map F=(F₁,…,Fₘ) has limit L=(L₁,…,Lₘ) exactly when every component has the matching scalar limit. Necessity follows from |Fⱼ−Lⱼ|≤‖F−L‖₂. For sufficiency, choose each component tolerance ε/√m and take the minimum of the corresponding finitely many positive δ values; the output norm is then less than ε.

The map F(x,y)=(x+y,x−y,xy) is continuous everywhere because its components are polynomials. At (1,2) its limit is (3,−1,2). A componentwise check retains all three outputs; silently plotting only one component answers a different question.

In scientific work, a scalar field, a vector field and a parameter-to-measurement map can have different input and output spaces. A spatial field at a fixed time, a time-dependent field and a list of measurements across trials are distinct domains. State which variables can vary and which are held fixed before interpreting a graph or an approach.

The next lesson studies rates of change and local linear approximations. The domain and continuity distinctions established here remain essential: an axis direction may leave the domain, and information from axis traces does not automatically control a full neighbourhood.

Use finite samples to test predictions, then return to the proof

Choose a named function to inspect a small exact input grid and its domain restrictions. Square-root and logarithm entries retain the exact radicand or positive argument; any decimal shown is explicitly a numerical display approximation. A missing grid value means the selected tuple is outside the stated domain, not that its limit has been computed.

For the three punctured quotients, compare the chosen line and parabola paths at nonzero t. Predict the symbolic path limit before reading the finite samples. The distinction between a grid, a path limit and an all-domain proof remains visible even when numerical values appear close.

Use the level-set descriptions, explicit origin extensions and vector-image tests to connect the displays to the article’s arguments. The controls are finite fixed examples, not a general theorem prover or an arbitrary expression parser. Preserve the chosen domains and codomains when drawing a conclusion.

Worked examples

Try it yourself

1. Does a vector-valued map assign several competing values to one input?

   Show the answer and reasoning

   No.

   It assigns one ordered output vector. Its components are coordinates of that one output.

2. Is (2,4) in the domain of ln(y−x²)?

   Show the answer and reasoning

   No.

   The argument is 0; a real logarithm requires strict positivity.

3. Is (9/5,12/5) in the domain of √(9−x²−y²), and what is the value?

   Show the answer and reasoning

   Yes, with value 0.

   The squared coordinates sum to 81/25+144/25=9, so the point is on the included boundary.

4. Where do the graph and image of a map R²→R³ naturally live?

   Show the answer and reasoning

   The graph is a subset of R⁵; the image is a subset of R³.

   A graph retains both input coordinates and all three output coordinates.

5. What is the level −1 of x²+2y²?

   Show the answer and reasoning

   The empty set.

   The formula is nonnegative for all real inputs.

6. What is the level 3 of √(9−x²−y²)?

   Show the answer and reasoning

   Only the origin.

   Its radicand must equal 9, forcing x²+y²=0.

7. For z=x²+2y², is the trace y=0 the same as the level z=0?

   Show the answer and reasoning

   No.

   The trace is the parabola z=x² in a vertical plane; the zero level in the input plane is only (0,0).

8. Can changing f(a) change a limit computed through punctured-domain points approaching a?

   Show the answer and reasoning

   No.

   The condition 0<‖x−a‖ excludes that one input, although its assigned value matters for continuity.

9. Must a limit at the boundary of the hemisphere disk use points outside the disk?

   Show the answer and reasoning

   No.

   Limits are relative to the actual real domain. Exterior points have no real value for that formula.

10. Do ten agreeing approach paths prove a joint limit?

    Show the answer and reasoning

    No.

    The definition requires one bound for all sufficiently close allowed points; a path not sampled may behave differently.

11. Can even all straight lines through a point have the same limit while the joint limit fails?

    Show the answer and reasoning

    Yes.

    The function x²y/(x⁴+y²) tends to 0 on every fixed line, but equals 1/2 on y=x² away from the origin.

12. If both iterated limits exist and agree, must the joint limit exist?

    Show the answer and reasoning

    No.

    For xy/(x²+y²) both are 0, while a diagonal path gives 1/2 and an axis path gives 0.

13. For |f(x,y)|≤4√(x²+y²) near the origin, give a δ that proves limit 0 for tolerance ε.

    Show the answer and reasoning

    If the bound holds for 0<ρ<r with r>0, take δ=min(r,ε/4).

    Then 0<ρ<δ keeps the input within the bound’s neighbourhood and gives |f|≤4ρ<ε.

14. For r=x²y/(x²+y²), what happens if the origin extension is assigned 1?

    Show the answer and reasoning

    It is discontinuous there.

    The punctured joint limit is 0, which does not equal the assigned value 1.

15. Can assigning a value at the origin make xy/(x²+y²) continuous there?

    Show the answer and reasoning

    No.

    Its path-dependent punctured values have no joint limit to match.

16. Are the three quotient examples continuous at every nonzero input?

    Show the answer and reasoning

    Yes.

    Their numerators and denominators are polynomials and their denominators are positive away from the origin.

17. Why does choosing component tolerance ε/√m suffice for a map with m real output coordinates?

    Show the answer and reasoning

    The Euclidean norm of the error is then less than ε.

    The sum of m squared component errors is less than m(ε²/m)=ε²; use the minimum of the finitely many δ values.

18. Can a function be continuous at an isolated domain point?

    Show the answer and reasoning

    Yes; it is automatically continuous there.

    A sufficiently small relative neighbourhood contains only that point. Our limit convention separately requires an accumulation point for a punctured limit.

What to carry forward

• Several coordinates form one ordered input, and a vector-valued map still assigns one output to each allowed input.
• A formula, its domain, codomain and image answer different questions. Algebraic cancellation does not automatically fill excluded inputs.
• Graphs retain inputs and outputs; images retain outputs. Level sets live in the input space and need not be smooth curves.
• Limits approach through the actual domain at accumulation points. Continuity is stated at domain points, including relative boundaries.
• The joint-limit quantifiers require one δ controlling every sufficiently close allowed input. Point values are excluded from the punctured limit.
• Different valid approach limits disprove a joint limit. Finite grids, agreeing iterated limits and even all fixed straight lines do not prove one.
• Algebraic limit laws require their hypotheses. At a vanishing denominator, use a new argument rather than the quotient rule for limits.
• A uniform estimate such as |r|≤‖(x,y)‖ proves a limit in all directions at once.
• A continuous extension must assign the existing finite punctured limit; no assignment can repair path dependence.
• Vector-valued limits are componentwise in finite dimensions. Units, scales and the chosen domain remain part of a scientific interpretation.