Continuity

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We investigate what continuity means for real-valued functions of several variables.

This section investigates what it means for real -valued functions of $n$ -variables to be “continuous” .

We begin with a series of definitions. We are used to “open intervals” such as $(1,3)$ , which represents the set of all $x$ such that $1<x<3$ , and “closed intervals” such as $[1,3]$ , which represents the set of all $x$ such that $1\leq x\leq 3$ . We need analogous definitions for open and closed sets in $\R ^n$ .

We give these definitions in general, for when one is working in $\R ^n$ :

An open ball $B$ in $\R ^n$ centered at $\vec {a} = \vector {a_1,a_2,\dots ,a_n}$ with radius $r$ is the set of all vectors $\vec {x}=\vector {x_1,x_2,\dots ,x_n}$ such that $|\vec {x}-\vec {a}| < r$ . In $\R ^2$ an open ball is often called an open disk.
A point $\vec {p}$ (denoted by a vector) in $S$ is an interior point of $S$ if there is an open ball $B$ centered at $\vec {p}$ that contains only points in $S$ . We can write this in symbols as $\vec {p}\in B\subseteq S$
Let $S$ be a set of points in $\R ^n$ . A point $\vec {p}$ (denoted by a vector) in $\R ^n$ is a boundary point of $S$ if all open balls centered at $\vec {p}$ contain both points in $S$ and points not in $S$ .
A set $O$ is open if every point in $O$ is an interior point.
A set $C$ is closed if it contains all of its boundary points.
A set $S$ is bounded if there is an open ball $B$ centered at the origin of radius $M$ such that $S\subseteq B.$ A set that is not bounded is unbounded.

Given a set $S$ , we denote the boundary of $S$ by $\partial S$ .

Consider a closed disk $D$ in $\R ^2$ . Describe $\partial D$ .

Since $\partial D$ is the boundary of a closed disk in $\R ^2$ , $\partial D$ is a disk a circle a ball a line .

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.

We’ve already found the domain of this function to be $D = \{(x,y): \answer [given]{x^2/9+y^2/4}\leq 1\}.$ This is the region bounded by the ellipse $\frac {x^2}9+\frac {y^2}4=1$ . Since the region includes the boundary (indicated by the use of “ $\leq$ ”), the set containsdoes not contain all of its boundary points and hence is closed. The region is boundedunbounded as a disk of radius $4$ , centered at the origin, contains $D$ .

Determine if the domain of $F(x,y) = \frac {1}{x-y}$ is open, closed, or neither, and if it is bounded.

As we cannot divide by $0$ , we find the domain to be $D = \left \{(x,y):x-y\neq \answer [given]{0}\right \}.$ In other words, the domain is the set of all points $(x,y)$ not on the line $y=x$ . For your viewing pleasure, we have included a graph:

Note how we can draw an open disk around any point in the domain that lies entirely inside the domain, and also note how the only boundary points of the domain are the points on the line $y=x$ . We conclude the domain is an open seta closed setneither open nor closed set . Moreover, the set is boundedunbounded .

Limits

On to the definition of a limit! Recall that for functions of a single variable, we say that $\lim _{x\to a} f(x) = L$ if the value of $f(x)$ can be made arbitrarily close to $L$ for all $x$ sufficiently close, but not equal to, $x=a$ .

This easily allows us to make a similar definition for functions of several variables.

Suppose that $F$ is a real-valued function of $n$ variables. Assume that the domain of $F$ contains a small $n$ -dimensional ball centered at a point $\vec {a}$ , except, possibly, the point $\vec {a}$ . The limit of $F$ as $\vec {x}$ approaches $\vec {a}$ is $L$ if the value of $F(\vec {x})$ can be made as close as one wishes to $L$ for all $\vec {x}$ sufficiently close, but not equal to, $\vec {a}$ .

When this occurs, we write $\lim _{\vec {x}\to \vec {a}} F(\vec {x}) = L$

Suppose that $F$ is a real-valued function of two variables, $\vec {x} = \vector {x,y}$ , and $\vec {a} = \vector {a,b}$ . Then the statement $\lim _{\vec {x}\to \vec {a}} F(\vec {x}) = L$ can be also written as $\lim _{\point {x,y}\to \point {a,b} }F\left (x,y\right ) = L$

Suppose that $F$ is a function of three variables, $\vec {x} = \vector {x,y,z}$ , and $\vec {a} = \vector {a,b,c}$ . Similarly, the statement $\lim _{\vec {x}\to \vec {a}} F(\vec {x}) = L$ can be written equivalently as

$\lim _{\point {x,y,z}\to \point {a,b,c}} F\left (x,y,z\right ) = L$

While the intuitive idea behind limits seems to remain unchanged, something interesting is worth observing. One of the most important ideas for limits of a function of a single variable is the notion of a sided limit. For functions of a single variable, there were really only two natural ways for $x$ to become close to $a$ ; we could take $x$ to approach the point $a$ from the left or the right. For instance, $\lim _{x\to a^-}f(x)$ tells us to consider the inputs $x<a$ only. In fact, there’s a theorem that guarantees that $\lim _{x\to a} f(x) = L$ if and only if $\lim _{x\to a^-}f(x) =L$ and $\lim _{x\to a^+}f(x) =L$ , meaning that the function must approach the same value as the input approaches $a$ from both the left and the right.

On the other hand, there are now infinitely many ways for, e.g., $(x,y)\to (a,b)$ ; we can approach along a straight line path parallel to the $x$ -axis or $y$ -axis, other straight line paths, or even other types of curves.

In order to check whether a limit exists, do we have to verify that the function tends to the same value along infinitely many different paths?

While this may seem problematic, there is some good news; many of the limit laws from before still do hold now.

Limit Laws Let $F$ and $G$ be real-valued functions of $n$ variables, and $b$ , $L$ and $M$ be real numbers, where $\lim _{\vec {x}\to \vec {a}}F(\vec {x}) = L \quad \text {and}\quad \lim _{\vec {x}\to \vec {a}} G(\vec {x}) = M.$ Let $\vec {x} = \vector {x_{1},x_{2},...,x_{n}}$ , and $\vec {a} = \vector {a_{1},a_{2},...,a_{2}}$ . We also assume that the domain of $F$ and the domain of $G$ both contain a small ball with the center at $\vec {a}$ , except possibly the point $\vec {a}$ .

Constant Law: $\lim _{\vec {x}\to \vec {a}} b = b$
Identity Law: $\lim _{\vec {x}\to \vec {a}} x_i = a_i$ , where $i=1,..., n$
Sum/Difference Law: $\lim _{\vec {x}\to \vec {a}}\big (F(\vec {x})\pm G(\vec {x})\big ) = L\pm M$
Scalar Multiple Law: $\lim _{\vec {x}\to \vec {a}} b\cdot F(\vec {x}) = bL$
Product Law: $\lim _{\vec {x}\to \vec {a}} \left (F(\vec {x})\cdot G(\vec {x})\right ) = LM$
Quotient Law: $\lim _{\vec {x}\to \vec {a}} \frac {F(\vec {x})}{G(\vec {x})} = \frac {L}{M}$ , if $M\neq 0$

In practice, this allows us to compute many limits in a similar fashion as before.

Compute $\lim _{(x,y)\to (1,2)} \frac {2x+4y}{x-3y}$ .

We show how the above properties are used quite explicitly. $\begin{align*} \lim_{(x,y)\to(1,2)} \frac{2x+4y}{x-3y} & = \frac{\lim_{(x,y)\to(1,2)}(2x+4y)}{\lim_{(x,y)\to(1,2)}(x-3y)} \textrm{ by the quotient law. } \\ &= \frac{\lim_{(x,y)\to(1,2)}2x+\lim_{(x,y)\to(1,2)} 4y}{\lim_{(x,y)\to(1,2)}x-\lim_{(x,y)\to(1,2)}3y} \textrm{ by the sum/difference law. } \\ &= \frac{2\left[\lim_{(x,y)\to(1,2)}x\right]+4\left[\lim_{(x,y)\to(1,2)} y\right]}{\left[\lim_{(x,y)\to(1,2)}x\right]-3\left[\lim_{(x,y)\to(1,2)}y\right]} \textrm{ by the scalar multiple law. } \\ &= \frac{2[1]+4[2]}{[1]-3[2]} \textrm{ by the identity law. } \\ &= -2 \end{align*}$

Essentially, the above laws allow us to evaluate limits by directly substituting values into the given function, provided the end result is a constant. Henceforth, when a limit can be evaluated by direct substitution, we will not show the details.

As it turns out, another old technique works well too.

Compute $\lim _{(x,y)\to (0,0)} \frac {x^3+xy^2}{x^2+y^2}$ .

To compute this limit, note that direct substitution leads us to the indeterminate form $\frac {0}{0}$ . $\begin{align*} \lim_{(x,y)\to(0,0)}\frac{x^3+xy^2}{x^2+y^2} &=\lim_{(x,y)\to(0,0)} \frac{\answer[given]{x}\left(x^2+y^2\right)}{\answer[given]{x^2+y^2}}\\ &=\lim_{(x,y)\to(0,0)} x\\ &=\answer[given]{0} \end{align*}$

What allows us to perform the cancellation of the common factors of $x^2+y^2$ ? Note that when determining whether a limit exists or not, we must look near the point $\point {0,0}$ , but not at the point $\point {0,0}$ . No matter how close a point $\point {x,y}$ is to the point $\point {0,0}$ , as long as $\point {x,y} \neq \point {0,0}$ , then $x^2+y^2 \neq 0$ . So, this cancellation is valid.

Limits exist when functions locally look like a smooth sheet.

When limits don’t exist

When dealing with functions of a single variable we often encounter a situation where the two one-sided limits are not equal, i.e. $\lim _{x\to a^+}f(x) \ne \lim _{x\to a^-}f(x)$ In that case, we say that $\lim _{x\to a}f(x)$ does not exist.

In $\R ^n$ when $n\ge 2$ there are infinite paths along which $\vec {x}$ might approach $\vec {a}$ .

If it is possible to arrive at different limiting values by approaching $\vec {a}$ along different paths, the limit does not exist.

This is analogous to the left and right hand limits of single variable functions not being equal, implying that the limit does not exist. 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. 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.

Show that $\lim _{(x,y)\to (0,0)} \frac {3xy}{x^2+y^2}$ does not exist by finding the limits along the lines $y=mx$ .

Evaluating $\lim _{(x,y)\to (0,0)} \frac {3xy}{x^2+y^2}$ along the lines $y=mx$ means replace all $y$ ’s with $mx$ and evaluating the resulting limit: $\begin{align*} \lim_{(x,mx)\to(0,0)} \frac{3x(\answer[given]{m x})}{x^2+(\answer[given]{mx})^2} &=\lim_{x\to0} \frac{3mx^2}{x^2(m^2+1)}\\ &= \lim_{x\to 0}\frac{3m}{m^2+1}\\ &= \answer[given]{\frac{3m}{m^2+1}}. \end{align*}$ While the limit exists for each choice of $m$ , we get a different limit for each choice of $m$ . Suppose $m=0$ , then: $\lim _{x\to 0}\frac {3m}{m^2+1} = \answer [given]{0}$ Now suppose that $m=1$ , then: $\lim _{x\to 0}\frac {3m}{m^2+1} = \answer [given]{3/2}$ Since we find differing limiting values when computing the limit along different paths, we must conclude that the limit does not exist. We finish by presenting you with a plot of $F$ :

Show that $\lim _{(x,y)\to (0,0)} \frac {6x^2y}{x^4+y^2}$

does not exist.

Evaluating $\lim _{(x,y)\to (0,0)} \frac {6x^2y}{x^4+y^2}$ along the lines $y=mx$ means replace all $y$ ’s with $mx$ and evaluating the resulting limit: $\begin{align*} \lim_{(x,mx)\to(0,0)} \frac{6x^2(mx)}{x^4+(mx)^2} &=\lim_{x\to0} \frac{6mx^3}{x^2(m^2+x^2)}\\ &= \lim_{x\to 0}\frac{6mx}{m^2+x^2}\\ &= {\frac{0}{m^2}}=\answer[given]{0} \end{align*}$ Since the limit is equal to zero for each choice of $m$ , you may jump to a conclusion...but, wait! We have only examined what is happening when paths are straight lines. But, there are many other paths along which $(x,y)$ can approach $(0,0)$ . For example, consider the curve $y=x^2$ . Compute the limiting value of $F(x,y)$ along this path: $\begin{align*} \lim_{(x,x^2)\to(0,0)} \frac{6x^2(x^2)}{x^4+(x^2)^2} &=\lim_{x\to0} \frac{6x^4}{2x^4}\\ &=\answer[given]{ 3} \end{align*}$ Since we find differing limiting values when computing the limit along different paths, we must conclude that the limit does not exist.

Here we see the function:

If one approaches the origin along any line, you see the limit is zero, by following the path on the surface. However, if one approaches the origin along a parabola, then we see the limit does not exist, as approaching along the parabola $y=x^2$ gives a limit of $3$ . Thus this is a case where the limit does not exist.

Limits don’t exist when the function makes a large jump, or when the surface is somehow pinched.

Continuity

Now we will use the idea of a limit to define continuity.

Let $F:D\to \R$ be a real-valued function of $n$ variables, defined on the set $D\subseteq \R ^n$ that contains an open ball $B$ centered at $\vec {a}$ . $F$ is continuous at $\vec {a}$ , if

$F(\vec {a})$ exists.
$\lim _{\vec {x}\to \vec {a}} F(\vec {x})$ exists.
$\lim _{\vec {x}\to \vec {a}} F(\vec {x}) = F(\vec {a}).$

$F$ is continuous on an open ball $B$ if $F$ is continuous at all points in $B$ .

True or false: If $F:\R ^2\to \R$ and $G:\R ^2\to \R$ are continuous functions on an open disk $B$ , then $F\pm G$ is continuous on $B$ .

True False

Take any point, say, $\vec {a}$ in $B$ . Since $F$ and $G$ are both continuous on $B$ , they are both continuous at $\vec {a}$ . This means that $\lim _{\vec {x}\to \vec {a}} F(\vec {x}) = F(\vec {a})$ and $\lim _{\vec {x}\to \vec {a}} G(\vec {x}) = G(\vec {a})$ . This implies that

$\lim _{\vec {x}\to \vec {a}} (F+G)(\vec {x})=\lim _{\vec {x}\to \vec {a}} (F(\vec {x})+G(\vec {x}))=$ ( by the sum law )
$=\lim _{\vec {x}\to \vec {a}} F(\vec {x})+\lim _{\vec {x}\to \vec {a}} G(\vec {x})= F(\vec {a})+G(\vec {a})=(F+G)(\vec {a})$

True or false: If $F:\R ^2\to \R$ and $G:\R ^2\to \R$ are continuous functions on an open disk $B$ , then $F/G$ is continuous on $B$ .

True False

The function $F/G$ may or may not be continuous, it depends on whether $G(x,y)=0$ . If $G(x,y)=0$ , then $F/G$ not continuous at that point.

Composition Limit Law Let $f:I\to \R$ be a continuous function on an interval $I\subseteq \R$ . Let $G:D\to \R$ , where $D\subseteq \R ^n$ , be a function whose range is contained in (or equal to) $I$ . Then $\lim _{\vec {x}\to \vec {a}} f( G(\vec {x})) = f(\lim _{\vec {x}\to \vec {a}}G(\vec {x}))$

Composition of Composite Functions Let $G:D\to \R$ , where $D\subseteq \R ^n$ , be continuous on an open disk $B$ , where the range of $G$ on $B$ is $I$ , and let $f$ be a single variable function that is continuous on $I$ . Then, the function $f\circ G$ is continuous on $B$ . In other words, $\lim _{\vec {x}\to \vec {a}} (f\circ G)(\vec {x}) =\lim _{\vec {x}\to \vec {a}}f(G(\vec {x}))=f(G(\vec {a})),$ for all points $\vec {a}$ in $B$ .

Show that the function $F(x,y) = \sin (x^2\cos (y))$ is continuous for all points in $\R ^2$ .

Let $F_1(x,y) = x^2.$ Since $y$ is not actually used in the function, and polynomials are continuousare not continuous , we conclude $F_1$ is continuous everywhere. A similar statement can be made about $F_2(x,y) = \cos (y).$ Setting $F_3=F_1\cdot F_2$ we obtain a continuous function from $\R ^2\to \R$ . Since sine is continuousis not continuous for all real values, the composition of sine with $F_3$ is continuous. Hence, $\sin (F_3(x,y)) = \sin (x^2\cos y)$ is continuous everywhere. We finish by presenting you with a plot of $F$ :

Let $F(x,y) = \begin {cases} \frac {\cos (y)\sin (x)}{x} & x\neq 0 \\ \cos (y) & x=0 \end {cases}$ Is $F$ continuous at $(0,0)$ ? Is $F$ continuous everywhere?

To determine if $F$ is continuous at $(0,0)$ , we need to compare $\lim _{(x,y)\to (0,0)} F(x,y)\quad \text {to}\quad F(0,0).$ Applying the definition of $F$ , we see that: $F(0,0) = \answer [given]{1}$ We now consider the limit $\lim _{(x,y)\to (0,0)}F(x,y).$ Substituting $0$ for $x$ and $y$ in $(\cos (y)\sin (x))/x$ returns the indeterminate form $\relax \boldsymbol {\tfrac {0}{0}}$ , so we need to do more work to evaluate this limit.

Consider two related limits:

$\begin{align*} \lim_{(x,y)\to(0,0)} \cos(y)\\ \lim_{(x,y)\to(0,0)} \frac{\sin(x)}{x}. \end{align*}$ The first limit does not contain $x$ , and since $\cos (y)$ is continuous, $\begin{align*} \lim_{(x,y)\to(0,0)}\cos(y) &=\lim_{y\to 0} \cos(y) \\ &=\answer[given]{1} \end{align*}$ The second limit does not contain $y$ . But we know $\begin{align*} \lim_{(x,y)\to(0,0)} \frac{\sin(x)}{x} &= \lim_{x\to 0} \frac{\sin(x)}{x} \\ &= \answer[given]{1}. \end{align*}$ Finally, we can apply the Product Law
$\begin{align*} \lim_{(x,y)\to(0,0)} \frac{\cos(y)\sin(x)}{x}&= \lim_{(x,y)\to(0,0)}(\cos(y))\left(\frac{\sin(x)}{x}\right) \\ &=\left( \lim_{(x,y)\to(0,0)} \cos(y)\right)\left( \lim_{(x,y)\to(0,0)} \frac{\sin(x)}{x}\right) \\ &=\answer[given]{1}\cdot \answer[given]{1}. \end{align*}$ We have found that $\lim _{(x,y)\to (0,0)} \frac {\cos (y)\sin (x)}{x} = F(0,0)$ , so $F$ is continuous at $(0,0)$ .

A similar analysis shows that $F$ is continuous at all points in $\mathbb {R}^2$ . As long as $x\neq 0$ , we can evaluate the limit directly; when $x=0$ , a similar analysis shows that the limit is $\cos y$ . Thus we can say that $F$ is continuous everywhere. We finish by presenting you with a plot of $F$ :