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 $(1,3)$, which represents the set of all $x$ such that $$, and “closed intervals” such as $[1,3]$, which represents the set of all $x$ such that $1\le x\le 3$. We need analogous definitions for open and closed sets in the $x$-$y$ plane.
An open disk $B$ in ${\mathbb{R}}^{2}$ centered at $({x}_{0},{y}_{0})$ with radius $r$ is the set of all points $(x,y)$ such that $$.
Let $S$ be a set of points in ${\mathbb{R}}^{2}$. 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 $S$.
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 $S$.
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 $M$, contains $S$. A set that is not bounded is unbounded.
Figure 13.2.1 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 13.2.1(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.
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.
SolutionThis domain of this function was found in Example 13.1.2 to be $D=\{(x,y)|\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}\le 1\}$, 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 “$\le $”), the set contains all of its boundary points and hence is closed. The region is bounded 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.
SolutionAs we cannot divide by 0, we find the domain to be $D=\{(x,y)|x-y\ne 0\}$. In other words, the domain is the set of all points $(x,y)$ not on the line $y=x$.
The domain is sketched in Figure 13.2.2. 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 set. The set is unbounded.
Recall a pseudo-definition of the limit of a function of one variable: “$\underset{x\to c}{lim}f(x)=L$” means that if $x$ is “really close” to $c$, then $f(x)$ is “really close” to $L$. A similar pseudo-definition holds for functions of two variables. We’ll say that
“$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}f(x,y)=L$”
means “if the point $(x,y)$ is really close to the point $({x}_{0},{y}_{0})$, then $f(x,y)$ is really close to $L$.” The formal definition is given below.
Let $S$ be an open set containing $({x}_{0},{y}_{0})$, and let $f$ be a function of two variables defined on $S$, except possibly at $({x}_{0},{y}_{0})$. The limit of $f(x,y)$ as $(x,y)$ approaches $({x}_{0},{y}_{0})$ is $L$, denoted
$$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}f(x,y)=L,$$ |
means that given any $\u03f5>0$, there exists $\delta >0$ such that for all $(x,y)\ne ({x}_{0},{y}_{0})$, if $(x,y)$ is in the open disk centered at $({x}_{0},{y}_{0})$ with radius $\delta $, then $$
The concept behind Definition 13.2.2 is sketched in Figure 13.2.3. Given $\u03f5>0$, find $\delta >0$ such that if $(x,y)$ is any point in the open disk centered at $({x}_{0},{y}_{0})$ in the $x$-$y$ plane with radius $\delta $, then $f(x,y)$ should be within $\u03f5$ of $L$.
Computing limits using this definition is rather cumbersome. The following theorem allows us to evaluate limits much more easily.
Let $b$, ${x}_{0}$, ${y}_{0}$, $L$ and $K$ be real numbers, let $n$ be a positive integer, and let $f$ and $g$ be functions with the following limits:
$$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}f(x,y)=L\mathit{\hspace{1em}}\text{and}\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}g(x,y)=K.$$ |
The following limits hold.
1. Constants:
$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}b=b$
2. Identity:
$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}x={x}_{0}$; $\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}y={y}_{0}$
3. Sums/Differences:
$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}\left(f(x,y)\pm g(x,y)\right)=L\pm K$
4. Scalar Multiples:
$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}b\cdot f(x,y)=bL$
5. Products:
$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}f(x,y)\cdot g(x,y)=LK$
6. Quotients:
$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}f(x,y)/g(x,y)=L/K$, ($K\ne 0$)
7. Powers:
$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}f{(x,y)}^{n}={L}^{n}$
8. Roots:
$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}\sqrt[n]{f(x,y)}=\sqrt[n]{L}$
(when $n$ is odd or $L\ge 0$)
This theorem can be proved by the same arguments as the analogous results for functions of one variable in Theorem 1.3.1. Combined with Theorems 1.3.3 and 1.3.4 of Section 1.3, this allows us to evaluate many limits.
Evaluate the following limits:
$$1.\underset{(x,y)\to (1,\pi )}{lim}\left(\frac{y}{x}+\mathrm{cos}(xy)\right)\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}2.\underset{(x,y)\to (0,0)}{lim}\frac{3xy}{{x}^{2}+{y}^{2}}$$ |
Solution
The aforementioned theorems allow us to simply evaluate $y/x+\mathrm{cos}(xy)$ when $x=1$ and $y=\pi $. If an indeterminate form is returned, we must do more work to evaluate the limit; otherwise, the result is the limit. Therefore
$\underset{(x,y)\to (1,\pi )}{lim}{\displaystyle \frac{y}{x}}+\mathrm{cos}(xy)$ | $={\displaystyle \frac{\pi}{1}}+\mathrm{cos}\pi $ | ||
$=\pi -1.$ |
We attempt to evaluate the limit by substituting 0 in for $x$ and $y$, but the result is the indeterminate form “$0/0$.” To evaluate this limit, we must “do more work,” but we have not yet learned what “kind” of work to do. Therefore we cannot yet evaluate this limit.
When dealing with functions of a single variable we also considered one-sided limits and stated
$$\underset{x\to c}{lim}f(x)=L\mathit{\hspace{1em}}\text{if and only if both}\mathit{\hspace{1em}}\underset{x\to {c}^{+}}{lim}f(x)=L\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\underset{x\to {c}^{-}}{lim}f(x)=L.$$ |
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})$. 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 any possible path. It is possible to arrive at different limiting values by approaching $({x}_{0},{y}_{0})$ along different paths. If this happens, we say that $\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}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.
Watch the video:
Showing a Limit Does Not Exist from https://youtu.be/q9xIdF33ql8
(a)
Show $\underset{(x,y)\to (0,0)}{lim}{\displaystyle \frac{3xy}{{x}^{2}+{y}^{2}}}$ does not exist by finding the limits along the lines $y=mx$.
(b)
Show $\underset{(x,y)\to (0,0)}{lim}{\displaystyle \frac{\mathrm{sin}(xy)}{x+y}}$ does not exist by finding the limit along the path $y=-\mathrm{sin}x$.
Solution
Evaluating $\underset{(x,y)\to (0,0)}{lim}{\displaystyle \frac{3xy}{{x}^{2}+{y}^{2}}}$ along the lines $y=mx$ means replace all $y$’s with $mx$ and evaluating the resulting limit:
$\underset{(x,mx)\to (0,0)}{lim}{\displaystyle \frac{3x(mx)}{{x}^{2}+{(mx)}^{2}}}$ | $=\underset{x\to 0}{lim}{\displaystyle \frac{3m{x}^{2}}{{x}^{2}({m}^{2}+1)}}$ | ||
$=\underset{x\to 0}{lim}{\displaystyle \frac{3m}{{m}^{2}+1}}$ | |||
$={\displaystyle \frac{3m}{{m}^{2}+1}}.$ |
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.
Let $f(x,y)=\frac{\mathrm{sin}(xy)}{x+y}$. We are to show that $\underset{(x,y)\to (0,0)}{lim}f(x,y)$ does not exist by finding the limit along the path $y=-\mathrm{sin}x$. First, however, consider the limits found along the lines $y=mx$ as done above.
$\underset{(x,mx)\to (0,0)}{lim}{\displaystyle \frac{\mathrm{sin}\left(x(mx)\right)}{x+mx}}$ | $=\underset{x\to 0}{lim}{\displaystyle \frac{\mathrm{sin}(m{x}^{2})}{x(m+1)}}$ | ||
$=\underset{x\to 0}{lim}{\displaystyle \frac{\mathrm{sin}(m{x}^{2})}{x}}\cdot {\displaystyle \frac{1}{m+1}}.$ |
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$, $\underset{(x,y)\to (0,0)}{lim}f(x,y)=0$. Now consider the limit along the path $y=-\mathrm{sin}x$:
$\underset{(x,-\mathrm{sin}x)\to (0,0)}{lim}{\displaystyle \frac{\mathrm{sin}\left(-x\mathrm{sin}x\right)}{x-\mathrm{sin}x}}$ | $=\underset{x\to 0}{lim}{\displaystyle \frac{\mathrm{sin}\left(-x\mathrm{sin}x\right)}{x-\mathrm{sin}x}}$ |
Now apply L’Hôpital’s Rule twice:
$=\underset{x\to 0}{lim}{\displaystyle \frac{\mathrm{cos}\left(-x\mathrm{sin}x\right)\left(-\mathrm{sin}x-x\mathrm{cos}x\right)}{1-\mathrm{cos}x}}\mathit{\hspace{1em}}\left(\text{\u201c}=0/0\text{\u201d}\right)$ | |||
$=\underset{x\to 0}{lim}{\displaystyle \frac{-\mathrm{sin}\left(-x\mathrm{sin}x\right){\left(-\mathrm{sin}x-x\mathrm{cos}x\right)}^{2}+\mathrm{cos}\left(-x\mathrm{sin}x\right)\left(-2\mathrm{cos}x+x\mathrm{sin}x\right)}{\mathrm{sin}x}}$ | |||
$=\text{\u201c}-2/0\text{\u201d}\Rightarrow \text{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=-\mathrm{sin}x$, which lies in the domain of $f(x,y)$ for all $x\ne 0$, the limit does not exist. Since the limit is not the same along every path to $(0,0)$, we say $\underset{(x,y)\to (0,0)}{lim}{\displaystyle \frac{\mathrm{sin}(xy)}{x+y}}$ does not exist.
Let $f(x,y)={\displaystyle \frac{5{x}^{2}{y}^{2}}{{x}^{2}+{y}^{2}}}$. Find $\underset{(x,y)\to (0,0)}{lim}f(x,y)$.
SolutionIt is relatively easy to show that along any line $y=mx$, the limit is 0. This is not enough to prove that the limit exists, as demonstrated in the previous example, but it tells us that if the limit does exist then it must be 0.
To prove the limit is 0, we apply Definition 13.2.2. Let $\u03f5>0$ be given. We want to find $\delta >0$ such that if $$, then $$.
Set $$. Note that $$ for all $(x,y)\ne (0,0)$, and that if $$, then $$.
Let $$. Consider $\left|f(x,y)-0\right|$:
$\left|f(x,y)-0\right|$ | $=\left|{\displaystyle \frac{5{x}^{2}{y}^{2}}{{x}^{2}+{y}^{2}}}-0\right|$ | ||
$=\left|{x}^{2}\cdot {\displaystyle \frac{5{y}^{2}}{{x}^{2}+{y}^{2}}}\right|$ | |||
$$ | |||
$$ | |||
$=\u03f5.$ |
Thus if $$ then $$, which is what we wanted to show. Thus $\underset{(x,y)\to (0,0)}{lim}{\displaystyle \frac{5{x}^{2}{y}^{2}}{{x}^{2}+{y}^{2}}}=0$.
We also have a multivariable version of the squeeze theorem.
Let $S$ be an open set containing $({x}_{0},{y}_{0})$. Suppose $f(x,y)$, $g(x,y)$, and $h(x,y)$ are defined on $S$ except possibly at $({x}_{0},{y}_{0})$ and both
$$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}g(x,y)=L\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}h(x,y)=L.$$ |
If $g(x,y)\le f(x,y)\le h(x,y)$ for all $(x,y)\in S$ except possibly at $({x}_{0},{y}_{0})$, then
$$\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}f(x,y)=L.$$ |
This theorem provides other proofs of the previous example.
We have
$$0\le \frac{5{x}^{2}{y}^{2}}{{x}^{2}+{y}^{2}}\le \frac{5{x}^{2}{y}^{2}+5{y}^{4}}{{x}^{2}+{y}^{2}}=\frac{5{y}^{2}({x}^{2}+{y}^{2})}{{x}^{2}+{y}^{2}}=5{y}^{2}.$$ |
Since $0\to 0$ and $5{y}^{2}\to 0$ as $(x,y)\to (0,0)$ we have
$$\underset{(x,y)\to (0,0)}{lim}\frac{5{x}^{2}{y}^{2}}{{x}^{2}+{y}^{2}}=0$$ |
by the Squeeze Theorem.
If we set $x=r\mathrm{cos}\theta $ and $y=r\mathrm{sin}\theta $ we have $(x,y)\to 0$ as $r\to 0$. Now
$$\frac{5{x}^{2}{y}^{2}}{{x}^{2}+{y}^{2}}=\frac{5{r}^{4}{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta}{{r}^{2}({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta )}=5{r}^{2}{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta .$$ |
Thus
$$0\le \frac{5{x}^{2}{y}^{2}}{{x}^{2}+{y}^{2}}=5{r}^{2}{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta \le 5{r}^{2}.$$ |
Since $5{r}^{2}\to 0$ as $r\to 0$ we have again
$$\underset{(x,y)\to (0,0)}{lim}\frac{5{x}^{2}{y}^{2}}{{x}^{2}+{y}^{2}}=0$$ |
by the Squeeze Theorem.
Definition 1.6.1 defines what it means for a function of one variable to be continuous. In brief, it meant that the function always equaled its limit. We define continuity for functions of two variables in a similar way as we did for functions of one variable.
Let a function $f(x,y)$ be defined on an open disk $B$ containing the point $({x}_{0},{y}_{0})$.
$f$ is continuous at $({x}_{0},{y}_{0})$ if $\underset{(x,y)\to ({x}_{0},{y}_{0})}{lim}f(x,y)=f({x}_{0},{y}_{0})$.
$f$ is continuous on an open set $S$ if $f$ is continuous at each points in $S$. (We say that $f$ is continuous everywhere if $f$ is continuous on ${\mathbb{R}}^{2}$.)
It follows that if $f$ is a continuous function of one variable, then $f(x)$ (or $f(y)$) is a continuous function of the variables $(x,y)$.
Let $f(x,y)=\{\begin{array}{cc}\frac{\mathrm{cos}y\mathrm{sin}x}{x}\hfill & x\ne 0\hfill \\ \mathrm{cos}y\hfill & x=0\hfill \end{array}$. Is $f$ continuous at $(0,0)$? Is $f$ continuous everywhere?
SolutionTo determine if $f$ is continuous at $(0,0)$, we need to compare $\underset{(x,y)\to (0,0)}{lim}f(x,y)$ to $f(0,0)$.
Applying the definition of $f$, we see that $f(0,0)=\mathrm{cos}0=1$.
We now consider the limit $\underset{(x,y)\to (0,0)}{lim}f(x,y)$. Substituting $0$ for $x$ and $y$ in $(\mathrm{cos}y\mathrm{sin}x)/x$ returns the indeterminate form “0/0”, so we need to do more work to evaluate this limit.
Consider two related limits: $\underset{(x,y)\to (0,0)}{lim}\mathrm{cos}y$ and $\underset{(x,y)\to (0,0)}{lim}{\displaystyle \frac{\mathrm{sin}x}{x}}$. The first limit does not contain $x$, and since $\mathrm{cos}y$ is continuous,
$$\underset{(x,y)\to (0,0)}{lim}\mathrm{cos}y=\underset{y\to 0}{lim}\mathrm{cos}y=\mathrm{cos}0=1.$$ |
The second limit does not contain $y$. By Theorem 1.3.6 we can say
$$\underset{(x,y)\to (0,0)}{lim}\frac{\mathrm{sin}x}{x}=\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1.$$ |
Finally, Theorem 13.2.1 of this section states that we can combine these two limits as follows:
$\underset{(x,y)\to (0,0)}{lim}{\displaystyle \frac{\mathrm{cos}y\mathrm{sin}x}{x}}$ | $=\underset{(x,y)\to (0,0)}{lim}(\mathrm{cos}y)\left({\displaystyle \frac{\mathrm{sin}x}{x}}\right)$ | ||
$=\left(\underset{(x,y)\to (0,0)}{lim}\mathrm{cos}y\right)\left(\underset{(x,y)\to (0,0)}{lim}{\displaystyle \frac{\mathrm{sin}x}{x}}\right)$ | |||
$=(1)(1)$ | |||
$=1.$ |
We have found that $\underset{(x,y)\to (0,0)}{lim}{\displaystyle \frac{\mathrm{cos}y\mathrm{sin}x}{x}}=f(0,0)$, 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 $x\ne 0$, we can evaluate the limit directly; when $x=0$, a similar analysis shows that the limit is $\mathrm{cos}y$. Thus we can say that $f$ is continuous everywhere. A graph of $f$ is given in Figure 13.2.4. Notice how it has no breaks, jumps, etc.
The following theorems are very similar to Theorems 1.6.1 and 1.6.2, giving us ways to combine continuous functions to create other continuous functions.
Let $f$ and $g$ be continuous on an open set $S$, let $c$ be a real number, and let $n$ be a positive integer. The following functions are continuous on $S$.
1. Sums/Differences:
$f\pm g$
2. Constant Multiples:
$c\cdot f$
3. Products:
$f\cdot g$
4. Quotients:
$f/g$ (as longs as $g\ne 0$ on $B$)
5. Powers:
${f}^{n}$
6. Roots:
$\sqrt[n]{f}$ (if $f\ge 0$ on $B$ or $n$ is odd)
Let $f$ be continuous on $S$, where the range of $f$ on $S$ is $J$, and let $g$ be a single variable function that is continuous on $J$. Then
$$(g\circ f)(x,y)=g(f(x,y)),$$ |
is continuous on $S$.
Let $f(x,y)=\mathrm{sin}({x}^{2}\mathrm{cos}y)$. Show $f$ is continuous everywhere.
SolutionWe will apply Theorems 1.6.1, 13.2.3, and 13.2.4. We let ${f}_{1}(x,y)={x}^{2}$. Since $y$ is not actually used in the function, and polynomials are continuous (by Theorem 1.6.1), we conclude ${f}_{1}$ is continuous everywhere. A similar statement can be made about ${f}_{2}(x,y)=\mathrm{cos}y$. Part 3 of Theorem 13.2.3 states that ${f}_{3}={f}_{1}\cdot {f}_{2}$ is continuous everywhere, and Theorem 13.2.4 states the composition of sine with ${f}_{3}$ is continuous: that is, $\mathrm{sin}({f}_{3})=\mathrm{sin}({x}^{2}\mathrm{cos}y)$ is continuous everywhere.
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 13.2.1 and 13.2.3 are not redefined but their analogous meanings should be clear to the reader.
(a)
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 $\sqrt{{(x-{x}_{0})}^{2}+{(y-{y}_{0})}^{2}+{(z-{z}_{0})}^{2}}=r$.
(b)
Let $D$ be an open set in ${\mathbb{R}}^{3}$ containing $({x}_{0},{y}_{0},{z}_{0})$, and let $f(x,y,z)$ be a function of three variables defined on $D$, except possibly at $({x}_{0},{y}_{0},{z}_{0})$. The limit of $f(x,y,z)$ as $(x,y,z)$ approaches $({x}_{0},{y}_{0},{z}_{0})$ is $L$, denoted
$$\underset{(x,y,z)\to ({x}_{0},{y}_{0},{z}_{0})}{lim}f(x,y,z)=L,$$
means that given any $\u03f5>0$, there is a $\delta >0$ such that for all $(x,y,z)\ne ({x}_{0},{y}_{0},{z}_{0})$, if $(x,y,z)$ is in the open ball centered at $({x}_{0},{y}_{0},{z}_{0})$ with radius $\delta $, then $$.
(c)
Let $f(x,y,z)$ be defined on an open ball $B$ containing $({x}_{0},{y}_{0},{z}_{0})$. Then $f$ is continuous at $({x}_{0},{y}_{0},{z}_{0})$ if $\underset{(x,y,z)\to ({x}_{0},{y}_{0},{z}_{0})}{lim}f(x,y,z)=f({x}_{0},{y}_{0},{z}_{0})$.
These definitions can also be extended naturally to apply to functions of four or more variables. Theorems 13.2.3 and 13.2.4 also applies to function of three or more variables, allowing us to say that the function
$$f(x,y,z)=\frac{{e}^{{x}^{2}+y}\sqrt{{y}^{2}+{z}^{2}+3}}{\mathrm{sin}(xyz)+5}$$ |
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 multivariable context.
Describe in your own words the difference between boundary and interior points of a set.
Give an example of a closed, bounded set.
Give an example of a closed, unbounded set.
Give an example of a open, bounded set.
Give an example of a open, unbounded set.
In Exercises 7–10., a set $S$ is given.
Give one boundary point and one interior point, when possible, of $S$.
State whether $S$ is open, closed, or neither.
State whether $S$ is bounded or unbounded.
$S=\left\{(x,y)\right|{\displaystyle \frac{{\left(x-1\right)}^{2}}{4}}+{\displaystyle \frac{{\left(y-3\right)}^{2}}{9}}\le 1\}$
$S=\left\{(x,y)\right|y\ne {x}^{2}\}$
$S=\left\{(x,y)\right|{x}^{2}+{y}^{2}=1\}$
$S=\left\{(x,y)\right|y>\mathrm{sin}x\}$
In Exercises 11–18.:
Find the domain $D$ of the given function.
State whether $D$ is an open or closed set.
State whether $D$ is bounded or unbounded.
$f(x,y)=\sqrt{9-{x}^{2}-{y}^{2}}$
$f(x,y)=\sqrt{y-{x}^{2}}$
$f(x,y)={\displaystyle \frac{1}{\sqrt{y-{x}^{2}}}}$
$f(x,y)={\displaystyle \frac{{x}^{2}-{y}^{2}}{{x}^{2}+{y}^{2}}}$
$f(x,y)={x}^{2}+3xy+{y}^{2}$
$f(x,y)={\displaystyle \frac{{x}^{2}+y+1}{x-2y+3}}$
$f(x,y)={\displaystyle \frac{1}{\sqrt{1-{x}^{2}-{y}^{2}}}}$
$f(x,y)=\mathrm{ln}\left(xy\right)$
In Exercises 19–24., a limit is given. Evaluate the limit along the paths given, then state why these results show the given limit does not exist.