Functions of Several Variables

A function of several variables assigns one output to each admissible input tuple. Its domain is a set of tuples, not merely a separate interval for each input. Coupled restrictions, level sets and approach paths all carry information that a one-dimensional graph cannot show.

Domain analysis comes before drawing a surface or calculating a derivative. A picture can suggest the geometry, but the defining conditions determine which points actually belong.

On this page

Specify inputs and output type

A scalar function can be written $f:D\subseteq\mathbb R^2\to\mathbb R$ . Each pair in $D$ receives one real output. The function $f(x,y)=x^2+2y^2$ is scalar-valued despite having two inputs.

A relation such as $z^2=x^2+y^2$ does not by itself assign a unique $z$ . Selecting the positive or negative square-root branch defines different functions. The branch and domain are part of the definition, not decorations added after a formula is chosen.

Intersect all restrictions

Consider

f(x,y)=\frac{\ln(9-x^2-y^2)}{y-1}.

The logarithm requires $x^2+y^2<9$ , while the denominator requires $y\ne1$ . Therefore

D=\{(x,y):x^2+y^2<9,\ y\ne1\}.

This is an open disk of radius three with the part of the line $y=1$ inside it removed. The separate inequalities $-3<x<3$ and $-3<y<3$ are necessary but insufficient. The point $(2.5,2.5)$ satisfies them but lies outside the disk.

Domain of ln(9−x²−y²)/(y−1)

Tinting marks the disk interior; excluded sets remain explicitly marked.

Diagram explanation and exact data

Separate coordinate bounds would include inadmissible corner points. The logarithm and denominator restrictions must both hold.

Domain of ln(9−x²−y²)/(y−1): exact reference data
Input	Status	Reason
(0,0)	included	positive log argument, nonzero denominator
(0,1)	excluded	zero denominator
(3,0)	excluded	log argument zero
(2.5,2.5)	excluded	outside disk

A square root permits zero radicand, while a logarithm does not. Check each operation separately. The several-variable function reference develops domain and graph representations.

Prove a range with bounds and witnesses

For

g(x,y)=\sqrt{9-x^2-y^2},

the domain is the closed disk $x^2+y^2\le9$ . The output satisfies $0\le g\le3$ . To establish that every value in this interval occurs, choose any $c\in[0,3]$ . The input $(\sqrt{9-c^2},0)$ belongs to the domain and gives $g=c$ .

Thus the range is exactly $[0,3]$ . An upper bound alone would not establish the whole range. The graph is the upper hemisphere of a sphere, including its circular boundary.

Read levels and traces differently

For $g(x,y)=c$ , with $0\le c\le3$ ,

x^2+y^2=9-c^2.

Levels below three are circles; level three is the origin. Values outside the range have empty level sets. A function of three variables can have level surfaces: $x^2+y^2+z^2=c$ gives spheres for positive $c$ , a point at zero and no real points for negative $c$ .

A trace instead fixes an input. For $f=x^2+2y^2$ , setting $y=b$ gives $x^2+2b^2$ . This vertical slice is different from a level curve, which fixes the output. A finite collection of slices does not uniquely determine every possible surface between them.

Define limits using domain points

The statement $f(x,y)\to L$ as $(x,y)\to(a,b)$ requires every sufficiently close admissible input, other than the point itself, to have output close to $L$ .

Formally, for each $\varepsilon>0$ , there is a $\delta>0$ such that

0<\sqrt{(x-a)^2+(y-b)^2}<\delta\quad\Longrightarrow\quad|f(x,y)-L|<\varepsilon

for points in the domain. The target is understood to be an accumulation point. At a boundary, approaches remain within the domain; values outside it must not be invented.

Continuity additionally requires that the limit equal the assigned function value. This relative-domain interpretation matters at boundaries and removed subsets.

Different path limits disprove a limit

For $f(x,y)=xy/(x^2+y^2)$ away from the origin, both axes give zero. Along $y=x$ , the function equals $1/2$ . The limit does not exist.

Two disagreeing paths give a counterexample to a common limit. Several agreeing paths provide evidence but not a proof. A limit assertion concerns every sufficiently close domain point, not just selected approaches.

Every straight line can still miss a curved approach

Take

h(x,y)=\frac{x^2y}{x^4+y^2},\qquad(x,y)\ne(0,0).

Along $y=mx$ , for fixed $m\ne0$ ,

h(x,mx)=\frac{mx}{x^2+m^2}\longrightarrow0.

Both coordinate axes also give zero. Along the curved approach $y=x^2$ , however,

h(x,x^2)=\frac{x^4}{2x^4}=\frac12

for nonzero $x$ . The multivariable limit fails despite agreement along every fixed straight line.

The line-by-line result does not supply a uniform bound over all approaches. A curved path can change its direction as it nears the point. This distinction prevents an infinite-looking collection of examples from being mistaken for a general argument.

Prove a limit with one bound for all directions

For

q(x,y)=\frac{x^2y}{x^2+y^2},\qquad(x,y)\ne(0,0),

we have

|q(x,y)|=|y|\frac{x^2}{x^2+y^2}\le|y|\le\sqrt{x^2+y^2}.

The last expression tends to zero independently of direction. Choosing $\delta=\varepsilon$ proves the required implication, so $q\to0$ . Defining $q(0,0)=0$ gives a continuous extension.

The inequality controls all nearby inputs at once. The multivariable limits reference explains the role of such estimates alongside path counterexamples.

Carry domain geometry into later calculus

Partial derivatives vary a single coordinate, while directional derivatives approach along a chosen line. Boundary points may not permit two-sided changes in every direction. Standard interior differentiability definitions assume an open neighbourhood; constrained domains need their own interpretation.

Integration bounds also depend on domain shape. A disk cannot be represented by independent constant bounds on both coordinates without including extra points. Algebraic simplification can preserve values on an original domain while leaving its excluded points excluded.

Write restrictions explicitly and carry them into transformations, derivatives and integrals. A visually smooth surface is not evidence that a denominator restriction has disappeared.

Practice questions

Find the real domain of $\ln(9-x^2-y^2)/(y-1)$ .
Find the range of $\sqrt{9-x^2-y^2}$ and justify attainability.
Does a zero limit along every straight line through the origin establish a zero two-variable limit?

Check your answers

The domain is $x^2+y^2<9$ with $y\ne1$ .
The range is $[0,3]$ . For each such $c$ , the input $(\sqrt{9-c^2},0)$ gives output $c$ .
No. The function $x^2y/(x^4+y^2)$ tends to zero along every fixed line but equals $1/2$ along $y=x^2$ .