A function of several variables takes a point as its input. The first job is to find which points are allowed. We can then study the function through its graph, level sets, and limits.

Domains and graphs

For a function $f:\mathbb R^2\to\mathbb R$, the domain is the set of pairs $(x,y)$ for which the formula makes sense. Its graph is the surface

$$z=f(x,y)$$

in $\mathbb R^3$.

Example: combine two domain restrictions

Consider

$$f(x,y)=\sqrt{1-x^2-y^2}+\ln\left(y+\frac12\right).$$

The square root requires

$$1-x^2-y^2\ge0 \quad\Longleftrightarrow\quad x^2+y^2\le1.$$

The logarithm requires a strictly positive input:

$$y+\frac12>0 \quad\Longleftrightarrow\quad y>-\frac12.$$

Both conditions must hold at once, so

$$\boxed{ \operatorname{Dom}(f) =\left\{(x,y)\in\mathbb R^2: x^2+y^2\le1\text{ and }y>-\frac12 \right\}. }$$

This is the part of the closed unit disk above $y=-1/2$. The circular boundary is included wherever it also satisfies the logarithm condition. The horizontal chord at $y=-1/2$ is excluded.

The domain is an intersection of two restrictions

Level curves and contour maps

A level curve of $f(x,y)$ is a set of inputs where the function has one fixed value:

$$f(x,y)=c.$$

On the graph $z=f(x,y)$, this comes from the horizontal trace at height $z=c$. The trace is a curve in the plane $z=c$; the level curve records the same $(x,y)$ coordinates in the input plane. For more background on traces, see Distance and Quadric Surfaces.

A collection of level curves is called a contour map. It shows how the output changes without requiring a three-dimensional surface drawing.

Example: the contours of $\sqrt{xy}$

Let

$$f(x,y)=\sqrt{xy}.$$

The square root is real when $xy\ge0$. Thus the domain is

$$(x\ge0\text{ and }y\ge0) \quad\text{or}\quad (x\le0\text{ and }y\le0).$$

It consists of quadrants I and III together with the coordinate axes.

Because a principal square root is nonnegative, a level value must satisfy $c\ge0$. The equation $f(x,y)=c$ becomes

$$\sqrt{xy}=c \quad\Longleftrightarrow\quad xy=c^2.$$

For the first few levels:

$$\begin{aligned} c=0&:\quad x=0\text{ or }y=0,\\ c=1&:\quad xy=1,\\ c=2&:\quad xy=4. \end{aligned}$$

For $c>0$, the level curve is a hyperbola with one branch in quadrant I and one in quadrant III. Larger values of $c$ lie farther from the axes.

Contour map for the square root of xy

Functions of three variables

For a function $f:\mathbb R^3\to\mathbb R$, the full graph

$$w=f(x,y,z)$$

lives in $\mathbb R^4$, so we cannot draw it directly in ordinary space. We can still study the three-dimensional level surfaces

$$f(x,y,z)=c.$$

For example, take

$$f(x,y,z)=1000-x^2-y^2-z^2.$$

The level equation is

$$x^2+y^2+z^2=1000-c.$$

This gives

$$\begin{aligned} c=0&:\quad x^2+y^2+z^2=1000, &&\text{a sphere of radius }\sqrt{1000},\\ c=500&:\quad x^2+y^2+z^2=500, &&\text{a sphere of radius }\sqrt{500},\\ c=1000&:\quad x^2+y^2+z^2=0, &&\text{the single point }(0,0,0). \end{aligned}$$

The last level set is a degenerate sphere of radius $0$. A level set need not be an ordinary surface; it can collapse to a point or even be empty.

From one-variable limits to two variables

For a one-variable function, the statement

$$\lim_{x\to a}f(x)=L$$

means that every requested output tolerance $\varepsilon>0$ can be met by choosing an input tolerance $\delta>0$ such that

$$0<|x-a|<\delta \quad\Longrightarrow\quad |f(x)-L|<\varepsilon.$$

In two variables, replace the one-dimensional distance $|x-a|$ with the Euclidean distance between $(x,y)$ and $(a,b)$:

$$\lVert(x,y)-(a,b)\rVert =\sqrt{(x-a)^2+(y-b)^2}.$$

Formal definition

Let $f:D\to\mathbb R$, where $D\subseteq\mathbb R^2$, and let $(a,b)$ be an accumulation point of $D$. We write

$$\lim_{(x,y)\to(a,b)}f(x,y)=L$$

if, for every $\varepsilon>0$, there is a $\delta>0$ such that every $(x,y)\in D$ satisfies

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

The input condition describes a punctured disk: points within radius $\delta$ of $(a,b)$, with the center removed. The output condition requires the corresponding heights to lie in the band

$$L-\varepsilon<f(x,y)<L+\varepsilon.$$

A punctured input disk controls an output band

The puncture matters. A limit depends on nearby values, not on the value at the center. The function may have a different value at $(a,b)$ or be undefined there.

A warning from a surface

The function

$$g(x,y)=\frac{x^2-y^2}{x^2+y^2}$$

is undefined at $(0,0)$. Its graph near the origin suggests that approaching from different directions may produce incompatible values, so a single limit should not be assumed from a picture. A proof of nonexistence must exhibit the contradiction formally; that argument comes after the definition-based example below.

Example: prove a limit from the definition

We will prove

$$\lim_{(x,y)\to(0,0)} \frac{3x^2y}{x^2+y^2}=0.$$

The expression is undefined at the origin, but the punctured-neighborhood condition excludes that point.

To determine how small $\delta$ must be for a given $\varepsilon$, first bound the error in terms of the distance $\sqrt{x^2+y^2}$. For $(x,y)\ne(0,0)$,

$$\begin{aligned} \left|\frac{3x^2y}{x^2+y^2}-0\right| &=3|y|\frac{x^2}{x^2+y^2}\\ &\le3|y|\\ &\le3\sqrt{x^2+y^2}. \end{aligned}$$

The first inequality uses $0\le x^2/(x^2+y^2)\le1$. The second uses $y^2\le x^2+y^2$.

Now let $\varepsilon>0$ and choose

$$\delta=\frac{\varepsilon}{3}.$$

If

$$0<\sqrt{x^2+y^2}<\delta,$$

then

$$\begin{aligned} \left|\frac{3x^2y}{x^2+y^2}-0\right| &\le3\sqrt{x^2+y^2}\\ &<3\delta\\ &=\varepsilon. \end{aligned}$$

This is exactly the implication required by the definition. Therefore,

$$\boxed{ \lim_{(x,y)\to(0,0)} \frac{3x^2y}{x^2+y^2}=0. }$$