Continuity

Learning Outcomes

State the conditions for continuity of a function of two variables.
Verify the continuity of a function of two variables at a point.
Calculate the limit of a function of three or more variables and verify the continuity of the function at a point.

In Continuity, we defined the continuity of a function of one variable and saw how it relied on the limit of a function of one variable. In particular, three conditions are necessary for $f\,(x)$ to be continuous at point $x=a$ :

$f\,(a)$ exists.
$\displaystyle{\lim_{x\to{a}}}f\,(x)$ exists.
$\displaystyle{\lim_{x\to{a}}}f\,(x)=f\,(a)$ .

These three conditions are necessary for continuity of a function of two variables as well.

Definition

A function $f\,(x,\ y)$ is continuous at a point $(a,\ b)$ in its domain if the following conditions are satisfied:

$f\,(a,\ b)$ exists.
$\displaystyle{\lim_{(x,\ y)\to{(a,\ b)}}}f\,(x,\ y)$ exists.
$\displaystyle{\lim_{(x,\ y)\to(a,\ b)}}f\,(x,\ y)=f\,(a,\ b)$ .

Example: Demonstrating Continuity for a Function of Two Variables

Show that the function $f\,(x,\ y)=\frac{3x+2y}{x+y+1}$ is continuous at point $(5,-3)$ .

Show Solution

There are three conditions to be satisfied, per the definition of continuity. In this example,

a = 5

$a=5$ and

b = - 3

$b=-3$ .

$f\,(a,\ b)$ exists. This is true because the domain of the function $f$ consists of those ordered pairs for which the denominator is nonzero (i.e., $x+y+1\neq{0}$ ). Point $(5,-3)$ satisfies this condition. Furthermore,
$f\,(a,\ b)=f\,(5,-3)=\frac{3(5)+2(-3)}{5+(-3)+1}=\frac{15-6}{2+1}=3.$
$\displaystyle\lim_{(x,\ y)\to(a,\ b)}f\,(x,\ y)$ exists. This is also true:
$\begin{array}{ccc}\hfill {\displaystyle\lim_{(x,\ y)\to(a,\ b)}f\,(x,\ y)} & =\hfill & {\displaystyle\lim_{(x,\ y)\to(5,\ -3)}\frac{3x+2y}{x+y+1}}\hfill \\ \hfill & =\hfill & {\frac{\displaystyle\lim_{(x,\ y)\to(5,\ -3)}(3x+2y)}{\displaystyle\lim_{(x,\ y)\to(5,\ -3)}(x+y+1)}}\hfill \\ \hfill & =\hfill & {\frac{15-6}{5-3+1}}\hfill \\ \hfill & =\hfill & {3.}\hfill \\ \hfill \end{array}$
$\displaystyle\lim_{(x,\ y)\to(a,\ b)}f\,(x,\ y)=f\,(a,\ b)$ . This is true because we have just shown that both sides of this equation equal three.

Try it

Show that the function $f\,(x,y)=\sqrt{26-2x^{2}-y^{2}}$ is continuous at point $(2,-3)$ .

Show Solution

The domain of $f$ contains the ordered pair $(2,-3)$ because $f\,(a,b)=f\,(2,-3)=\sqrt{16-2(2)^{2}-(-3)^{2}}=3$
$\displaystyle\lim_{(x,y)\to(a,b)}f\,(x,y)=3$
$\displaystyle\lim_{(x,y)\to(a,b)}f\,(x,y)=f\,(a,b)=3$

Continuity of a function of any number of variables can also be defined in terms of delta and epsilon. A function of two variables is continuous at a point $(x_0,y_0)$ in its domain for every $\epsilon>0$ there exists a $\delta>0$ such that, whenever $\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$ it is true, $|f(x,y)-f(a,b)|<\epsilon$ . This definition can be combined with the formal definition (that is, the epsilon-delta definition) of continuity of a function of one variable to prove the following theorems:

the sum of continuous functions is continuous

If $f(x,y)$ is continuous at $(x_0,y_0)$ , and $g(x,y)$ is continuous at $(x_0,y_0)$ , then $f(x,y)+g(x,y)$ is continuous at $(x_0,y_0)$ .

the product of continuous functions is continuous

If $g(x)$ is continuous at $x_0$ , and $h(y)$ is continuous at $y_0$ , then $f(x,y)=g(x)h(y)$ is continuous at $(x_0,y_0)$ .

the composition of continuous functions is continuous

Let $g$ be a function of two variables from a domain $D\subseteq\mathbb{R}^{2}$ to a range $R\subseteq\mathbb{R}$ . Suppose $g$ is continuous at some point $(x_0,y_0)\in{D}$ and define $z_0=g(x_0,y_0)$ . Let $f$ be a function that maps $\mathbb{R}$ to $\mathbb{R}$ such that $z_0$ is in the domain of $f$ . Last, assume $f$ is continuous at $z_0$ . Then $f\circ{g}$ is continuous at $(x_0,y_0)$ as shown in the following figure.

A shape is shown labeled the domain of g with point (x, y) inside of it. From the domain of g there is an arrow marked g pointing to the range of g, which is a straight line with point z on it. The range of g is also marked the domain of f. Then there is another arrow marked f from this shape to a line marked range of f.

Figure 1. The composition of two continuous functions is continuous.

Let’s now use the previous theorems to show continuity of functions in the following examples.

Example: More Examples of Continuity of a Function of Two Variables

Show that the functions $f\,(x,\ y)=4x^{3}y^{2}$ and $g\,(x,\ y)=\cos{(4x^{3}y^{2})}$ are continuous everywhere.

Show Solution

The polynomials $g\,(x)=4x^{3}$ and $h\,(y)=y^{2}$ are continuous at every real number, and therefore by the product of continuous functions theorem, $f\,(x,\ y)=4x^{3}y^{2}$ is continuous at every point $(x,\ y)$ in the $xy$ -plane. Since $f\,(x,\ y)=4x^{3}y^{2}$ is continuous at every point $(x,\ y)$ in the $xy$ -plane and $g\,(x)=\cos{x}$ is continuous at every real number $x$ , the continuity of the composition of functions tells us that $g\,(x,\ y)=\cos{(4x^{3}y^{2})}$ is continuous at every point $(x,\ y)$ in the $xy$ -plane.

Try it

Show that the functions $f\,(x,\ y)=2x^{2}y^{3}+3$ and $g\,(x,\ y)=(2x^{2}y^{3}+3)^{4}$ are continuous everywhere.

Show Solution

The polynomials $g\,(x)=2x^{2}$ and $h\,(y)=y^{3}$ are continuous at every real number; therefore, by the product of continuous functions theorem, $f\,(x,\ y)=2x^{2}y^{3}$ is continuous at every point $(x,\ y)$ in the $xy$ -plane. Furthermore, any constant function is continuous everywhere, so $g\,(x,\ y)=3$ is continuous at every point $(x,\ y)$ in the $xy$ -plane. Therefore, $f\,(x,\ y)=2x^{2}y^{3}$ is continuous at every point $(x,\ y)$ in the $xy$ -plane. Last, $h\,(x)=x^{4}$ is continuous at every real number $x$ , so by the continuity of composite functions theorem $g\,(x,\ y)=(2x^{2}y^{3}+3)^{4}$ is continuous at every point $(x,\ y)$ in the $xy$ -plane.

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 4.10” here (opens in new window).

Functions of Three or More Variables

The limit of a function of three or more variables occurs readily in applications. For example, suppose we have a function $f\,(x,\ y,\ z)$ that gives the temperature at a physical location $(x,\ y,\ z)$ in three dimensions. Or perhaps a function $g\,(x,\ y,\ z,\ t)$ can indicate air pressure at a location $(x,\ y,\ z)$ at time $t$ . How can we take a limit at a point in $\mathbb{R}^{3}$ ? What does it mean to be continuous at a point in four dimensions?

The answers to these questions rely on extending the concept of a $\delta$ disk into more than two dimensions. Then, the ideas of the limit of a function of three or more variables and the continuity of a function of three or more variables are very similar to the definitions given earlier for a function of two variables.

Definition

Let $(x_0,\ y_0,\ z_0)$ be a point in $\mathbb{R}^{3}$ . Then, a $\delta$ ball in three dimensions consists of all points in $\mathbb{R}^{3}$ lying at a distance of less than $\delta$ from $(x_0,\ y_0,\ z_0)$ – that is,

{(x, y, z) \in R^{3} | \sqrt{(x - x_{0})^{2} + (y - y_{0})^{2} + (z - z_{0})^{2}} < δ}

$\large{\big\{(x,\ y,\ z)\in\mathbb{R}^{3}\big|\sqrt{(x-x_0)^{2}+(y-y_0)^{2}+(z-z_0)^{2}}\,<\delta\big\}}$ .

To define a $\delta$ ball in higher dimensions, add additional terms under the radical to correspond to each additional dimension. For example, given a point $P=(w_0,\ x_0,\ y_0,\ z_0)$ in $\mathbb{R}^{4}$ , a $\delta$ ball around $P$ can be described by

{(w, x, y, z) \in R^{4} | \sqrt{(w - w_{0})^{2} + (x - x_{0})^{2} + (y - y_{0})^{2} + (z - z_{0})^{2}} < δ}

$\large{\big\{(w,\ x,\ y,\ z)\in\mathbb{R}^{4}\big|\sqrt{(w-w_0)^{2}+(x-x_0)^{2}+(y-y_0)^{2}+(z-z_0)^{2}}\,<\delta\big\}}$ .

To show that a limit of a function of three variables exists at a point $(x_0,\ y_0,\ z_0)$ , it suffices to show that for any point in a $\delta$ ball centered at $(x_0,\ y_0,\ z_0)$ , the value of the function at that point is arbitrarily close to a fixed value (the limit value). All the limit laws for functions of two variables hold for functions of more than two variables as well.

Example: Finding the Limit of a Function of Three Variables

Find $\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}\frac{x^{2}y-3z}{2x+5y-z}.$

Show Solution

Before we can apply the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, the identity law, and the constant law,

\begin{array}{ccc} lim_{(x, y, z) \to (4, 1, - 3)} (2 x + 5 y - z) & = & 2 (lim_{(x, y, z) \to (4, 1, - 3)} x) + 5 (lim_{(x, y, z) \to (4, 1, - 3)} y) - (lim_{(x, y, z) \to (4, 1, - 3)} z) \\ = & 2 (4) + 5 (1) - (- 3) \\ = & 16. \end{array}

$\begin{array}{ccc}\hfill {\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}(2x+5y-z)} & =\hfill & {2\left(\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}x\right)+5\left(\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}y\right)-\left(\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}z\right)} \hfill \\ \hfill &=\hfill & {2(4)+5(1)-(-3)} \hfill \\ \hfill & =\hfill & {16.}\hfill \\ \hfill \end{array}$

Since this is nonzero, we next find the limit of the numerator. Using the product law, difference law, constant multiple law, and identity law,

\begin{array}{ccc} lim_{(x, y, z) \to (4, 1, - 3)} (x^{2} y - 3 z) & = & {(lim_{(x, y, z) \to (4, 1, - 3)} x)}^{2} (lim_{(x, y, z) \to (4, 1, - 3)} y) - 3 lim_{(x, y, z) \to (4, 1, - 3)} z \\ = & (4^{2}) (1) - 3 (- 3) \\ = & 16 + 9 \\ = & 25. \end{array}

$\begin{array}{ccc}\hfill {\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}(x^{2}y-3z)} & =\hfill & {\left(\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}x\right)^{2}\,\left(\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}y\right)-3\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}z}\hfill \\ \hfill & =\hfill & {(4^{2})(1)-3(-3)}\hfill \\ \hfill & =\hfill & {16+9}\hfill \\ \hfill & =\hfill & {25.}\hfill \\ \hfill \end{array}$

Last, applying the quotient law:

\begin{array}{ccc} lim_{(x, y, z) \to (4, 1, - 3)} \frac{x^{2} y - 3 z}{2 x + 5 y - z} & = & \frac{lim_{(x, y, z) \to (4, 1, - 3)} (x^{2} y - 3 z)}{lim_{(x, y, z) \to (4, 1, - 3)} (2 x + 5 y - z)} \\ = & \frac{25}{16} . \end{array}

$\begin{array}{ccc}\hfill {\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}\frac{x^{2}y-3z}{2x+5y-z}} &=\hfill & {\frac{\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}(x^{2}y-3z)}{\displaystyle\lim_{(x,\ y,\ z)\to(4,\ 1,\ -3)}(2x+5y-z)}}\hfill \\ \hfill & =\hfill & {\frac{25}{16}.} \hfill \\ \hfill \end{array}$

Try it

Find $\displaystyle\lim_{(x,\ y,\ z)\to(4,\ -1,\ 3)}\sqrt{13-x^{2}-2y^{2}+z^{2}}$ .

Show Solution

lim_{(x, y, z) \to (4, - 1, 3)} \sqrt{13 - x^{2} - 2 y^{2} + z^{2}} = 2

$\displaystyle\lim_{(x,\ y,\ z)\to(4,\ -1,\ 3)}\sqrt{13-x^{2}-2y^{2}+z^{2}}=2$

Module 4: Differentiation of Functions of Several Variables

Learning Outcomes

Definition

Example: Demonstrating Continuity for a Function of Two Variables

Try it

the sum of continuous functions is continuous

the product of continuous functions is continuous

the composition of continuous functions is continuous

Example: More Examples of Continuity of a Function of Two Variables

Try it

Functions of Three or More Variables

Definition

Example: Finding the Limit of a Function of Three Variables

Try it

Candela Citations