Continuity

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We investigate what continuity means for functions of several variables.

This section investigates what it means for functions $F:\R ^n\to \R$ 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 $P$ in $S$ is an interior point of $S$ if there is an open ball $B$ centered at $P$ that contains only points in $S$ . We can write this in symbols as $P\in B\subseteq S$
Let $S$ be a set of points in $\R ^n$ . A point $P$ in $\R ^n$ is a boundary point of $S$ if all open balls centered at $P$ contain both points in $S$ and points not in $S$ .
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 open ball $B$ centered at the origin of radius $M$ such that $S\subseteq B.$ A set that is not bounded is 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.

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 $D = \{(x,y):x-y\neq \answer [given]{0}\}.$ 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!

Suppose that $F:\R ^n\to \R$ , intuitively,

the limit of $F$ as $\vec {x}$ approaches $\vec {a}$ is $L$ ,

written $\lim _{\vec {x}\to \vec {a}} F(\vec {x}) = L,$ 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, $\vec {a}$ .

Suppose that $F:\R ^2\to \R$ , $\vec {x} = \vector {x,y}$ , and $\vec {a} = \vector {a,b}$ . What do we write in this case for $\lim _{\vec {x}\to \vec {a}} F(\vec {x}) = L$ ? $\lim _{\vector {\answer {x},\answer {y}}\to \vector {\answer {a},\answer {b}}} F(\answer {x},\answer {y}) = L$

Suppose that $F:\R ^3\to \R$ , $\vec {x} = \vector {x,y,z}$ , and $\vec {a} = \vector {a,b,c}$ . What do we write in this case for $\lim _{\vec {x}\to \vec {a}} F(\vec {x}) = L$ ? $\lim _{\vector {\answer {x},\answer {y},\answer {z}}\to \vector {\answer {a},\answer {b},\answer {c}}} F(\answer {x},\answer {y},\answer {z}) = L$

Compute: $\lim _{\vector {x,y}\to \vector {9,3}} \frac {x^2y-xy^3}{x^2-y^4}$

To compute this limit, we proceed by factoring. Write with me, $\begin{align*} \lim_{\vector{x,y}\to\vector{9,3}} \frac{x^2y-xy^3}{x^2-y^4} &=\lim_{\vector{x,y}\to\vector{9,3}} \frac{xy\left(\answer[given]{x-y^2}\right)}{(x+y^2)\left(\answer[given]{x-y^2}\right)}\\ &=\lim_{\vector{x,y}\to\vector{9,3}} \frac{xy}{(x+y^2)}\\ &=\answer[given]{\frac{3}{2}} \end{align*}$

Continuity

Now we will use our definition of a limit to define continuity.

Let $F:\R ^n\to \R$ be defined on an open disk $B$ centered at $\vec {a}$ .

$F$ is continuous at $\vec {a}$ , if $\lim _{\vec {x}\to \vec {a}} F(\vec {x}) = F(\vec {a}).$
$F$ is continuous on an open disk $B$ if $F$ is continuous at all points in $B$ .

To really use this definition, we need limit laws which in some sense are really continuity laws.

Limit Laws Let $F:\R ^n\to \R$ and $G:\R ^n\to \R$ be functions of several variables, $b$ , $L$ and $M$ are real numbers, $\begin{align*} \vec{x} &= \vector{x_1,x_2,\dots,x_n}\\ \vec{a} &= \vector{a_1,a_2,\dots,a_n}, \end{align*}$ where $\lim _{\vec {x}\to \vec {a}}F(\vec {x}) = L \quad \text {and}\quad \lim _{\vec {x}\to \vec {a}} G(\vec {x}) = M.$

Constant Law: $\lim _{\vec {x}\to \vec {a}} b = b$ .
Identity Law: $\lim _{\vec {x}\to \vec {a}} x_i = a_i$ .
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 $M\neq 0$ .

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

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 $f:\R \to \R$ be a continuous function on an interval $J$ . Let $G:\R ^n\to \R$ be a function whose range is contained in (or equal to) $J$ , 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 $G:\R ^n\to \R$ be continuous on an open disk $B$ , where the range of $G$ on $B$ is $J$ , and let $g$ be a single variable function that is continuous on $J$ . Then $g\circ F(\vec {x}) =g(F(\vec {x})),$ is continuous on $B$ .

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 $\lim _{\vector {x,y}\to \vector {0,0}} F(x,y)\quad \text {to}\quad F(0,0).$ Applying the definition of $F$ , we see that $F(0,0) = \answer [given]{1}$ We now consider the limit $\lim _{\vector {x,y}\to \vector {0,0}}F(x,y).$ 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:

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

When limits don’t exist

When dealing with functions of a single variable we often considered one-sided limits and stated $\lim _{x\to a}f(x) = L$ if and only if $\lim _{x\to a^+}f(x) =L \quad \textbf {and}\quad \lim _{x\to a^-}f(x) =L.$ That is, the limit is $L$ if and only if $F$ approaches $L$ when $x$ approaches $a$ from either direction.

In $\R ^n$ when $n>2$ there are infinite paths from which $\vec {x}$ might approach $\vec {a}$ . Now we have a fact, $\lim _{\vec {x}\to \vec {a}}F(\vec {x}) = L$ if and only if $F(\vec {x})\to L \quad {as}\quad \vec {x}\to \vec {a}$ along every path.

If it is possible to arrive at different limiting values by approaching 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. 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.

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

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

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

Let $F(x,y) = \frac {\sin (xy)}{x+y}.$ We will show that $\lim _{\vector {x,y}\to \vector {0,0}} F(x,y)$ does not exist by finding the limit along the path $y=-\sin (x)$ . First, however, let’s try the same trick that worked before and see what happens. Consider the limits found along the lines $y=mx$ as done above. $\begin{align*} \lim_{\vector{x,mx}\to\vector{0,0}} \frac{\sin\big(x(\answer[given]{mx})\big)}{x+\answer[given]{mx}} &= \lim_{x\to 0} \frac{\sin (mx^2)}{x(m+1)} \\ &= \lim_{x\to 0} \frac{\sin(mx^2)}{x}\cdot\answer[given]{\frac{1}{m+1}}. \end{align*}$ By applying L’Hôpital’s Rule, we can show this limit is $0$ except when $m=-1$ , that is, along the line $y=-x$ . This line is not in the domain of $F$ , so we have found the following fact: along every line $y=mx$ in the domain of $F$ , $\lim _{\vector {x,y}\to \vector {0,0}} F(x,y)=0.$ Now consider the limit along the path $y=-\sin (x)$ : $\begin{align*} \lim_{\vector{x,-\sin(x)}\to\vector{0,0}} \frac{\sin\big(-x\sin(x)\big)}{x-\sin(x)} &= \lim_{x\to0} \frac{\sin\big(-x\sin(x)\big)}{x-\sin(x)} \end{align*}$ Now apply L’Hôpital’s Rule twice to find a limit is of the form $\relax \boldsymbol {\tfrac {\#}{0}}$ . Hence the limit does not exist. Step back and consider what we have just discovered.

Along any line $y=mx$ in the domain of the $F(x,y)$ , the limit is $0$ .
However, along the path $y=-\sin (x)$ , which lies in the domain of the $F(x,y)$ for all $x\neq 0$ , the limit does not exist.

Since the limit is not the same along every path to $(0,0)$ , we say $\lim _{\vector {x,y}\to \vector {0,0}}\frac {\sin (xy)}{x+y}$ does not exist. We finish by presenting you with a plot of $F$ :