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 \(x_1, x_2, \cdots, x_n\):

\[f: \mathbb{R}^n \longrightarrow \mathbb{R}.\]

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 \[\lim_{(x,y) \to (a,b)}f(x,y) = \alpha\] 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 \(\varepsilon > 0\), there exists a \(\delta > 0\) such that, for all \((x,y) \in S\), if \((x,y) \in N_{\delta}(a,b)\) and \((x,y) \neq (a,b)\), then \(|f(x,y) - \alpha| < \varepsilon\).

or, in a logical form, \[\forall \varepsilon > 0 ~ \exists \delta > 0 ~ \forall (x,y)\in S ~ \left[(x,y) \in N_{\delta}((a,b))\land (x,y) \neq (a,b) \implies |f(x,y) - \alpha| < \varepsilon\right].\]

Theorem (Properties of limits)

  Let \(f(x,y)\) and \(g(x,y)\) be functions such that \[\begin{eqnarray} \lim_{(x,y)\to(a,b)}f(x,y) &=& \alpha,\\ \lim_{(x,y)\to(a,b)}g(x,y) &=& \beta. \end{eqnarray}\] Then, the following hold:

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

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

Example (Eg:polynomial). Clearly, \[\begin{eqnarray} \lim_{(x,y)\to (a,b)}x &=& a,\\ \lim_{(x,y)\to (a,b)}y &=& b. \end{eqnarray}\] 

Let \(f(x,y)\) be an arbitrary polynomial of \(x\) and \(y\). Then, \[\lim_{(x,y)\to (a,b)}f(x,y) = f(a,b).\] For example, if \(f(x,y) = x^3 + 2xy^2\), then \[\lim_{(x,y) \to (a,b)}f(x,y) = a^3 + 2ab^2.\] 

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 \(g(a,b) \neq 0\), we have

\[\lim_{(x,y) \to (a,b)}\frac{f(x,y)}{g(x,y)} = \frac{f(a,b)}{g(a,b)}.\]

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

\[\lim_{(x,y) \to (1,-1)}\frac{f(x,y)}{g(x,y)} = \frac{f(1,-1)}{g(1,-1)} = \frac{3}{1} = 3.\]

□

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 \(x\neq 0\), 

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

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 \(f(x,y) = \frac{xy^2}{x^2 + y^2}\) 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 \((x,y)\neq (0,0)\) implies \(r > 0\) (thus the above expression is always valid). Thus,

\[|f(x,y) - 0| = r|\cos\theta\sin^2\theta| =r|\cos\theta||\sin^2\theta| \leq r.\]

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

\[\lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2 + y^2} = 0.\] □

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

\[\lim_{(x,y)\to (a,b)}f(x,y) = f(a,b),\]

or, more formally,

\[\forall \varepsilon > 0 ~ \exists \delta > 0 ~ \forall (x,y)\in S ~ \left[(x,y) \in N_{\delta}((a,b))\implies |f(x,y) - f(a,b)| < \varepsilon\right].\]

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 \(g(a,b) \neq 0\).

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

\[\lim_{(x,y)\to(a,b)}f(x,y) = f(a,b).\]

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 \(g(a,b) \neq 0\).

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

\[f(x,y) = \left\{ \begin{array}{lc} \frac{xy^2}{x^2 + y^2} & ((x,y) \neq (0,0)),\\ 0 & ((x,y) = (0,0)). \end{array} \right.\]

For \((x,y) \neq (0,0)\), \(f(x,y)\) 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 \(f(a)\neq f(b)\) 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?