# Section12.2Limits and Continuity of Multivariable Functions¶ permalink

We continue with the pattern we have established in this text: after defining a new kind of function, we apply calculus ideas to it. The previous section defined functions of two and three variables; this section investigates what it means for these functions to be “continuous.”

We begin with a series of definitions. We are used to “open intervals” such as $(1,3)\text{,}$ which represents the set of all $x$ such that $1\lt x\lt 3\text{,}$ and “closed intervals” such as $[1,3]\text{,}$ which represents the set of all $x$ such that $1\leq x\leq 3\text{.}$ We need analogous definitions for open and closed sets in the $x$-$y$ plane.

##### Definition12.2.1Open Disk, Boundary and Interior Points, Open and Closed Sets, Bounded Sets

An open disk $B$ in $\mathbb{R}^2$ centered at $(x_0,y_0)$ with radius $r$ is the set of all points $(x,y)$ such that $\ds\sqrt{(x-x_0)^2+(y-y_0)^2} \lt r\text{.}$

Let $S$ be a set of points in $\mathbb{R}^2\text{.}$ A point $P$ in $\mathbb{R}^2$ is a boundary point of $S$ if all open disks centered at $P$ contain both points in $S$ and points not in $S\text{.}$

A point $P$ in $S$ is an interior point of $S$ if there is an open disk centered at $P$ that contains only points in $S\text{.}$

A set $S$ is open if every point in $S$ is an interior point.

A set $S$ is closed if it contains all of its boundary points.

A set $S$ is bounded if there is an $M>0$ such that the open disk, centered at the origin with radius $M\text{,}$ contains $S\text{.}$ A set that is not bounded is unbounded.

Figure 12.2.2 shows several sets in the $x$-$y$ plane. In each set, point $P_1$ lies on the boundary of the set as all open disks centered there contain both points in, and not in, the set. In contrast, point $P_2$ is an interior point for there is an open disk centered there that lies entirely within the set.

The set depicted in Figure 12.2.2(a) is a closed set as it contains all of its boundary points. The set in (b) is open, for all of its points are interior points (or, equivalently, it does not contain any of its boundary points). The set in (c) is neither open nor closed as it contains some of its boundary points.

##### Example12.2.3Determining open/closed, bounded/unbounded

Determine if the domain of the function $f(x,y)=\sqrt{1-\frac{x^2}9-\frac{y^2}4}$ is open, closed, or neither, and if it is bounded.

Solution
##### Example12.2.4Determining open/closed, bounded/unbounded

Determine if the domain of $f(x,y) = \frac1{x-y}$ is open, closed, or neither.

Solution

# Subsection12.2.1Limits

Recall a pseudo–definition of the limit of a function of one variable: “$\lim\limits_{x\to c}f(x) = L$” means that if $x$ is “really close” to $c\text{,}$ then $f(x)$ is “really close” to $L\text{.}$ A similar pseudo–definition holds for functions of two variables. We'll say that

“$\lim\limits_{(x,y)\to (x_0,y_0)} f(x,y) = L$”

means “if the point $(x,y)$ is really close to the point $(x_0,y_0)\text{,}$ then $f(x,y)$ is really close to $L\text{.}$” The formal definition is given below.

##### Definition12.2.6Limit of a Function of Two Variables

Let $S$ be an open set containing $(x_0,y_0)\text{,}$ and let $f$ be a function of two variables defined on $S\text{,}$ except possibly at $(x_0,y_0)\text{.}$ The limit of $f(x,y)$ as $(x,y)$ approaches $(x_0,y_0)$ is $L\text{,}$ denoted \begin{equation*} \lim\limits_{(x,y)\to (x_0,y_0)} f(x,y) = L, \end{equation*} means that given any $\varepsilon>0\text{,}$ there exists $\delta>0$ such that for all $(x,y)\neq (x_0,y_0)\text{,}$ if $(x,y)$ is in the open disk centered at $(x_0,y_0)$ with radius $\delta\text{,}$ then $\abs{f(x,y) - L}\lt \varepsilon\text{.}$

The concept behind Definition 12.2.6 is sketched in Figure 12.2.7. Given $\varepsilon>0\text{,}$ find $\delta>0$ such that if $(x,y)$ is any point in the open disk centered at $(x_0,y_0)$ in the $x$-$y$ plane with radius $\delta\text{,}$ then $f(x,y)$ should be within $\varepsilon$ of $L\text{.}$

Computing limits using this definition is rather cumbersome. The following theorem allows us to evaluate limits much more easily.

This theorem, combined with Theorems 1.3.3 and 1.3.5 of Section 1.3, allows us to evaluate many limits.

##### Example12.2.9Evaluating a limit

Evaluate the following limits: \begin{equation*} 1. \lim_{(x,y)\to (1,\pi)} \frac yx + \cos(xy) \qquad\qquad 2. \lim_{(x,y)\to (0,0)} \frac{3xy}{x^2+y^2} \end{equation*}

Solution

When dealing with functions of a single variable we also considered one–sided limits and stated \begin{equation*} \lim_{x\to c}f(x) = L \text{ if, and only if, } \lim_{x\to c^+}f(x) =L \textbf{ and} \lim_{x\to c^-}f(x) =L. \end{equation*}

That is, the limit is $L$ if and only if $f(x)$ approaches $L$ when $x$ approaches $c$ from either direction, the left or the right.

In the plane, there are infinite directions from which $(x,y)$ might approach $(x_0,y_0)\text{.}$ In fact, we do not have to restrict ourselves to approaching $(x_0,y_0)$ from a particular direction, but rather we can approach that point along a path that is not a straight line. It is possible to arrive at different limiting values by approaching $(x_0,y_0)$ along different paths. If this happens, we say that $\lim\limits_{(x,y)\to(x_0,y_0) } f(x,y)$ does not exist (this is analogous to the left and right hand limits of single variable functions not being equal).

Our theorems tell us that we can evaluate most limits quite simply, without worrying about paths. When indeterminate forms arise, the limit may or may not exist. If it does exist, it can be difficult to prove this as we need to show the same limiting value is obtained regardless of the path chosen. The case where the limit does not exist is often easier to deal with, for we can often pick two paths along which the limit is different.

##### Example12.2.10Showing limits do not exist
1. Show $\lim\limits_{(x,y)\to (0,0)} \frac{3xy}{x^2+y^2}$ does not exist by finding the limits along the lines $y=mx\text{.}$

2. Show $\lim\limits_{(x,y)\to (0,0)} \frac{\sin(xy)}{x+y}$ does not exist by finding the limit along the path $y=-\sin(x)\text{.}$

Solution
##### Example12.2.11Finding a limit

Let $\ds f(x,y) = \frac{5x^2y^2}{x^2+y^2}\text{.}$ Find $\lim\limits_{(x,y)\to (0,0)} f(x,y)\text{.}$

Solution

# Subsection12.2.2Continuity

Definition 1.5.1 defines what it means for a function of one variable to be continuous. In brief, it meant that the graph of the function did not have breaks, holes, jumps, etc. We define continuity for functions of two variables in a similar way as we did for functions of one variable.

##### Definition12.2.12Continuous

Let a function $f(x,y)$ be defined on an open disk $B$ containing the point $(x_0,y_0)\text{.}$

1. $f$ is continuous at $(x_0,y_0)$ if $\lim\limits_{(x,y)\to(x_0,y_0)} f(x,y) = f(x_0,y_0)\text{.}$

2. $f$ is continuous on $B$ if $f$ is continuous at all points in $B\text{.}$ If $f$ is continuous at all points in $\mathbb{R}^2\text{,}$ we say that $f$ is continuous everywhere.

##### Example12.2.13Continuity of a function of two variables

Let $\ds f(x,y) = \left\{ \begin{array}{rl} \frac{\cos(y) \sin(x) }{x} \amp x\neq 0 \\ \cos(y) \amp x=0 \end{array} \right.\text{.}$ Is $f$ continuous at $(0,0)\text{?}$ Is $f$ continuous everywhere?

Solution

The following theorem is very similar to Theorem 1.5.8, giving us ways to combine continuous functions to create other continuous functions.

##### Example12.2.16Establishing continuity of a function

Let $f(x,y) = \sin(x^2\cos(y) )\text{.}$ Show $f$ is continuous everywhere.

Solution

# Subsection12.2.3Functions of Three Variables

The definitions and theorems given in this section can be extended in a natural way to definitions and theorems about functions of three (or more) variables. We cover the key concepts here; some terms from Definitions 12.2.1 and 12.2.12 are not redefined but their analogous meanings should be clear to the reader.

##### Definition12.2.17Open Balls, Limit, Continuous
1. An open ball in $\mathbb{R}^3$ centered at $(x_0,y_0,z_0)$ with radius $r$ is the set of all points $(x,y,z)$ such that $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2} = r\text{.}$

2. Let $D$ be an open set in $\mathbb{R}^3$ containing $(x_0,y_0,z_0)\text{,}$ and let $f(x,y,z)$ be a function of three variables defined on $D\text{,}$ except possibly at $(x_0,y_0,z_0)\text{.}$ The limit of $f(x,y,z)$ as $(x,y,z)$ approaches $(x_0,y_0,z_0)$ is $L\text{,}$ denoted \begin{equation*} \lim_{(x,y,z)\to (x_0,y_0,z_0)} f(x,y,z) = L, \end{equation*} means that given any $\varepsilon >0\text{,}$ there is a $\delta >0$ such that for all $(x,y,z)\neq(x_0,y_0,z_0)\text{,}$ if $(x,y,z)$ is in the open ball centered at $(x_0,y_0,z_0)$ with radius $\delta\text{,}$ then $\abs{f(x,y,z) - L}\lt \varepsilon\text{.}$

3. Let $f(x,y,z)$ be defined on an open ball $B$ containing $(x_0,y_0,z_0)\text{.}$ $f$ is continuous at $(x_0,y_0,z_0)$ if $\lim\limits_{(x,y,z)\to (x_0,y_0,z_0)} f(x,y,z) = f(x_0,y_0,z_0)\text{.}$

These definitions can also be extended naturally to apply to functions of four or more variables. Theorem 12.2.15 also applies to function of three or more variables, allowing us to say that the function \begin{equation*} \ds f(x,y,z) = \frac{e^{x^2+y}\sqrt{y^2+z^2+3}}{\sin(xyz)+5} \end{equation*} is continuous everywhere.

When considering single variable functions, we studied limits, then continuity, then the derivative. In our current study of multivariable functions, we have studied limits and continuity. In the next section we study derivation, which takes on a slight twist as we are in a multivarible context.

# Subsection12.2.4Exercises

Terms and Concepts

In the following exercises, a set $S$ is given.

In the following exercises:

1. Find the domain $D$ of the given function.

2. State whether $D$ is an open or closed set.

3. State whether $D$ is bounded or unbounded.

In the following exercises, a limit is given. Evaluate the limit along the paths given, then state why these results show the given limit does not exist.