Multivariate functions: limits and continuity

We now consider the functions of the form $$y=f(x_1,x_2,\cdots ,x_n),$$ that is, functions multiple independent variables $x_1,x_2,\cdots ,x_n$:

$$f:{\mathbb{R}}^n⟶\mathbb{R}.$$

An $n$-variable function like $f(x_1,x_2,\cdots ,x_n)$ has a subset of ${\mathbb{R}}^n=(\mathbb{R}\times \mathbb{R}\times \cdots \times \mathbb{R})$ as its domain. In particular, we study the meaning of limits and continuity in high-dimensional spaces.

Remark. More generally, we may consider a function with its codomain in ${\mathbb{R}}^m$. □

In the following, we mainly work with functions on ${\mathbb{R}}^2$ rather than more general ${\mathbb{R}}^n$. There is a huge gap between functions on $\mathbb{R}$ and those on ${\mathbb{R}}^2$. Compared to this gap, the conceptual difference between functions on ${\mathbb{R}}^2$ and those on ${\mathbb{R}}^n,n>2$ is relatively small. Therefore, for most cases, understanding functions on ${\mathbb{R}}^2$ is (mostly) sufficient.

The above figure shows the graph of $f(x,y)=x^2+y^2$. Generally, the graph of a two-variable function is a surface, as in this example.

We often apply the following procedure to understand the graph of a two-variable (or bivariate) function. First, fix the value of $x$ to a constant, say, $x=a$. Then consider the cross-section of $x=a$ (a plane) and $z=f(x,y)$, which is represented as $z=f(a,y)$. Now, $z=f(a,y)$ is a function of only $y$, which can be plotted on the $y$-$z$ plane. For $f(x,y)=x^2+y^2$, we have $f(a,y)=a^2+y^2$ which is a parabola. Then, we change the value of $a$ and repeat the same process. We can apply the same procedure by fixing the value of $y$ instead. Each curve in the above figure can be obtained by this procedure, and the result is the mesh representation of the surface.

Limit of multivariate functions

Let $f(x,y)$ be a function on $S\subset {\mathbb{R}}^2$, and $(a,b)\in {\mathbb{R}}^2$ a point. Let's consider the notion of limit of $f(x,y)$ as the point $(x,y)$ approaches $(a,b)$.

To begin with, what do we mean by "$(x,y)$ approaches $(a,b)$"? Clearly, there are infinitely many ways $(x,y)$ can approach $(a,b)$: along a straight line, in a spiral, in an arbitrary curve, etc. When we consider the limit of $f(x,y)$, we need to consider all possible ways of approaching it.

Intuitively, the limit of $f(x,y)$ as $(x,y)$ approaches $(a,b)$ can be defined as follows:

• If $f(x,y)$ approaches a constant value $\alpha \in \mathbb{R}$ as $(x,y)$ approaches $(a,b)$, irrespectively of the way the approach is made, then we say $f(x,y)$ converges to $\alpha$ as $(x,y)$ approaches $(a,b)$, and denote $$\underset{(x,y)\to (a,b)}{lim}f(x,y)=\alpha$$ or $$f(x,y)\to \alpha \text{ as }(x,y)\to (a,b).$$

More formally, the definition is given as:

Definition (Limit of a function)

The function $f(x,y)$ on $S\subset {\mathbb{R}}^2$ is said to converge to $\alpha$ as $(x,y)\to (a,b)$ if the following condition holds:

• For all $\epsilon >0$, there exists a $\delta >0$ such that, for all $(x,y)\in S$, if $(x,y)\in N_{\delta }(a,b)$ and $(x,y)\ne (a,b)$, then $|f(x,y)-\alpha |<\epsilon$.

or, in a logical form, $$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta >0\mathrm{\forall }(x,y)\in S[(x,y)\in N_{\delta }((a,b))\wedge (x,y)\ne (a,b)⟹|f(x,y)-\alpha |<\epsilon ].$$

Theorem (Properties of limits)

  Let $f(x,y)$ and $g(x,y)$ be functions such that $$\begin{array}{rcl}\underset{(x,y)\to (a,b)}{lim}f(x,y)& =& \alpha ,\\ \underset{(x,y)\to (a,b)}{lim}g(x,y)& =& \beta .\end{array}$$ Then, the following hold:

1. For any $k,l\in \mathbb{R}$, $$\underset{(x,y)\to (a,b)}{lim}\{kf(x,y)+lg(x,y)\}=k\alpha +l\beta .$$
2. $$\underset{(x,y)\to (a,b)}{lim}\{f(x,y)g(x,y)\}=\alpha \beta .$$
3. If $\beta \ne 0$, $$\underset{(x,y)\to (a,b)}{lim}\frac{f(x,y)}{g(x,y)}=\frac{\alpha }{\beta }.$$

Proof. Exercise (similar to the one-variable version). ■

Example (Eg:polynomial). Clearly, $$\begin{array}{rcl}\underset{(x,y)\to (a,b)}{lim}x& =& a,\\ \underset{(x,y)\to (a,b)}{lim}y& =& b.\end{array}$$ 

Let $f(x,y)$ be an arbitrary polynomial of $x$ and $y$. Then, $$\underset{(x,y)\to (a,b)}{lim}f(x,y)=f(a,b).$$ For example, if $f(x,y)=x^3+2x{y}^2$, then $$\underset{(x,y)\to (a,b)}{lim}f(x,y)=a^3+2a{b}^2.$$ 

For the rational function $\frac{f(x,y)}{g(x,y)}$ of $x$ and $y$ where $f(x,y)$ and $g(x,y)$ are polynomials of $x$ and $y$, if $g(a,b)\ne 0$, we have

$$\underset{(x,y)\to (a,b)}{lim}\frac{f(x,y)}{g(x,y)}=\frac{f(a,b)}{g(a,b)}.$$

For example, if $f(x,y)=x^3+2x{y}^2$ and $g(x,y)=5x^2+xy+3y^3$, then

$$\underset{(x,y)\to (1,-1)}{lim}\frac{f(x,y)}{g(x,y)}=\frac{f(1,-1)}{g(1,-1)}=\frac{3}{1}=3.$$

□

Example. The function $f(x,y)=\frac{x^2-y^2}{x^2+y^2}$ does not have a limit as $(x,y)\to (0,0)$. To see this, consider a line $l:y=mx$ and move $(x,y)$ towards $(0,0)$ along this line. Then, if $x\ne 0$, 

$$f(x,mx)=\frac{x^2-m^2{x}^2}{x^2+m^2{x}^2}=\frac{1-m^2}{1+m^2}$$

which converges to $\frac{1-m^2}{1+m^2}$ as $x\to 0$. However, this limit depends on the slope $m$. For example, if $m=1$, the limit is 0; if $m=0$, the limit is 1. Thus, the "limit" of $f(x,y)$ depends on the way how $(x,y)$ approaches $(0,0)$. Hence, $f(x,y)$ has no limit at $(0,0)$. □

Example (Eg 1). Let us find the limit of $f(x,y)=\frac{x{y}^2}{x^2+y^2}$ as $(x,y)\to (0,0)$. Using the polar form $(x,y)=(r\cos \theta ,r\sin \theta )$, we have

$$f(r\cos \theta ,r\sin \theta )=\frac{r^3\cos \theta {\sin }^2\theta }{r^2({\cos }^2\theta +{\sin }^2\theta )}=r\cos \theta {\sin }^2\theta .$$

Note that $(x,y)\ne (0,0)$ implies $r>0$ (thus the above expression is always valid). Thus,

$$|f(x,y)-0|=r|\cos \theta {\sin }^2\theta |=r|\cos \theta ||{\sin }^2\theta |\le r.$$

As $(x,y)\to (0,0)$, $r\to 0$ irrespectively of the way of approaching. Therefore, we have

$$\underset{(x,y)\to (0,0)}{lim}\frac{x{y}^2}{x^2+y^2}=0.$$ □

Continuity of multivariate functions

Definition (Continuous function [multivariate])

The function $f(x,y)$ on $S\subset {\mathbb{R}}^2$ is said to be continuous at $(a,b)$ if

$$\underset{(x,y)\to (a,b)}{lim}f(x,y)=f(a,b),$$

or, more formally,

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta >0\mathrm{\forall }(x,y)\in S[(x,y)\in N_{\delta }((a,b))⟹|f(x,y)-f(a,b)|<\epsilon ].$$

Theorem (Properties of continuous functions)

Let $f(x,y)$ and $g(x,y)$ be functions that are continuous at $(x,y)=(a,b)$. Then the following functions are also continuous at $(x,y)=(a,b)$:

1. $kf(x,y)+lg(x,y)$ where $k,l\in \mathbb{R}$ are arbitrary constants.
2. $f(x,y)g(x,y)$.
3. $\frac{f(x,y)}{g(x,y)}$ if $g(a,b)\ne 0$.

Proof. Exercise (see the above Theorem (Properties of limits)). ■

Example. As we have seen on the above example (Eg:polynomial), for any polynomial function $f(x,y)$ and any point $(a,b)\in {\mathbb{R}}^2$, we have

$$\underset{(x,y)\to (a,b)}{lim}f(x,y)=f(a,b).$$

Therefore, all polynomial function are continuous everywhere in ${\mathbb{R}}^2$. Also, any rational function \frac{f(x,y)}{g(x,y)} (with $f,g$ being polynomial functions) is continuous at any $(a,b)\in {\mathbb{R}}^2$ provided that $g(a,b)\ne 0$.

Example. Is the following function continuous on ${\mathbb{R}}^2$?

$$f(x,y)=\{\begin{array}{lc}\frac{x{y}^2}{x^2+y^2}& ((x,y)\ne (0,0)),\\ 0& ((x,y)=(0,0)).\end{array}$$

For $(x,y)\ne (0,0)$, $f(x,y)$ is a rational function and its denominator is non-zero. Therefore, it is continuous on ${\mathbb{R}}^2\setminus \{(0,0)\}$. In the above Example (Eg 1), we have seen that \(\lim_{(x,y)\to(0,0)\}f(x,y) = 0\). But $f(0,0)=0$ by the definition above. Therefore, $f(x,y)$ is continuous at $(0,0)$. Hence, $f(x,y)$ is continuous everywhere on ${\mathbb{R}}^2$.

Finally, we provide the following theorems without proof.

Theorem (Intermediate Value Theorem)

Let $f(x)$ be a continuous function on a path-connected set $D$ in ${\mathbb{R}}^n$ such that $f(a)\ne f(b)$ for $a,b\in D$. Then, for any value $l$ between $f(a)$ and $f(b)$, there exists at least one $c\in D$ such that $f(c)=l$.

Exercise. Why should the domain $D$ be path-connected?

Theorem (Extreme Value Theorem)

The continuous function $f(x)$ on a bounded closed set $F\subset {\mathbb{R}}^n$ has maximum and minimum values. That is, there exist $c,d\in F$ such that $f(c)$ and $f(d)$ are the maximum and minimum values, respectively, of $f(x)$.

Exercise. Why should the domain $F$ be a bounded closed set?