## Limits

What does $\lim_{x\to x_0}f(x)=L$ mean? It means $f(x)$ can be made arbitrarily close to $L$, if $x$ is sufficiently close to (but different from) $x_0$. In other words, $\lim_{x\to x_0}f(x)=L$ means if you give me any positive number $\epsilon>0$, no matter how small, I can choose a number $\delta>0$ such that if the distance between $x\ (\neq x_0)$ and $x_0$ is less than $\delta$, then the distance between $f(x)$ and $L$ will be less than $\epsilon$.

Similarly, the notation $\lim_{(x,y)\to(x_0,y_0)}f(x,y)=L$ means we can get $f(x,y)$ as close to $L$ as we want, if we choose $(x,y)$ sufficiently close, but not equal to, $(x_0,y_0)$. Recall that in 1-space, the distance between $x$ and $x_0$ is given by $|x-x_0|$; in 2-space, the distance between two points $(x,y)$ and $(x_0,y_0)$ is given by $|(x,y)-(x_0,y_0)|=\sqrt{(x-x_0)^2+(y-y_0)^2}$. We express the informal idea for the limit of functions of two variables more precisely as follows.

Definition: Let $f:U\rightarrow \mathbb{R}$, where $U\subseteq \mathbb{R}^2$. We say $f(x,y)$ approaches $L$ as $(x,y)$ approaches $(x_0,y_0)$ and write
$$\lim_{(x,y)\rightarrow(x_0,y_0)} f(x,y)=L \quad (\text{or } f(x,y)\rightarrow L, \text{as } (x,y)\rightarrow (x_0,y_0))$$
if

1. every neighborhood of $(x_0,y_0)$ contains at least one member of $U$ other than $(x_0,y_0)$,*
2. given any number $\epsilon>0$, there exists a number $\delta>0$ such that $|f(x,y)-L|<\epsilon$ holds for any $(x,y)$ in $U$ which satisfies $0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$.

* Equivalently, we can say every neighborhood of $(x_0,y_0)$ contains an infinite number of members of $U$.

In the above definition:

• Condition (a) guarantees that $(x_0,y_0)$ is not an isolated point. An isolated point is a point with a neighborhood that contains no other member of the domain.
• Condition (a) lets us define the limit of the function at a boundary point of $U$.
• $(x_0,y_0)$ may or may not be in $U$.
• Another notation for the limit defined above is:
$$\lim_{{x\rightarrow x_0}\atop{y\rightarrow y_0}} f(x,y)=L$$
Example 1
Using Definition 1 ($\epsilon-\delta$ definition), show
$$\lim_{(x,y)\rightarrow (x_0,y_0)} x=x_0.$$
Solution
Assume $\epsilon>0$ is given. We want to determine $\delta>0$ such that if
$0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta,\quad \text{then}\quad |x_0-x|<\epsilon.$ We note that
\begin{align*}
|x-x_0|=\sqrt{(x-x_0)^2}\leq \sqrt{(x-x_0)^2+(y-y_0)^2}<\delta.
\end{align*}
So if we choose $\delta\leq \epsilon$, automatically we will have $|x-x_0|<\epsilon$. Therefore, $\lim_{(x,y)\rightarrow (x_0,y_0)} x=x_0$.
Example 2
Using the definition of a limit ($\epsilon-\delta$ definition) to show
$$\lim_{(x,y)\to(0,0)}\frac{x^2 y^2}{x^2+y^2}=0.$$

Solution

Let $(x,y)$ satisfy $0<\sqrt{(x-0)^2+(y-0)^2}<\delta$. Because $(x,y)\neq (0,0)$, we can simplify $|f(x,y)-0$ as

\begin{align*}\left|\frac{x^2y^2}{x^2+y^2}-0\right|=&|x^2|\left|\frac{y^2}{x^2+y^2}\right|
\leq |x^2|\leq |x^2+y^2|<\delta^2
\end{align*}
Therefore, if we choose $\delta<\sqrt{\epsilon}$, then
$$0<\sqrt{(x-0)^2+(y-0)^2}<\delta \Rightarrow |f(x,y)-0|=\left|\frac{x^2y^2}{x^2+y^2}\right|<\epsilon$$

Note that the function is not defined at $(0,0)$.

Graph of $f$ is shown below

Example 3
Given the function
$$f(x,y)=\frac{\sin(x^2+y^2)}{x^2+y^2}$$
find $\lim_{(x,y)\rightarrow (0,0)} f(x,y)$ if it exists.

Solution
Let's use the polar coordinates:
$x=r\cos\theta,\quad y=r\sin\theta.$ Then $x^2+y^2=r^2$ and
$f(x,y)=f(r,\theta)=\frac{\sin r^2}{r^2}$ For simplicity we call $r^2=u$ and write $f$ as $\sin u/u$. In the calculus of a single variable, we learned that $\lim_{u\rightarrow 0}\frac{\sin u}{u}=1$. So we conclude $\lim_{(x,y)\rightarrow (0,0)} f(x,y)=1$.

Again note that the function is not defined at $(0,0)$.

Graph of $f$ is shown below.

As the definitions of limits in the one-dimensional case and scalar fields are analogous, many familiar properties of limits can be extended to functions of several variables.

Theorem: If $\lim_{(x,y)\rightarrow (x_0,y_0)}f(x,y)=L$, $\lim_{(x,y)\rightarrow (x_0,y_0)}g(x,y)=M$, and $c$ is a real number then

1. $\lim_{(x,y)\rightarrow (x_0,y_0)}(f(x,y)\pm g(x,y))=L\pm M$

2. $\lim_{(x,y)\rightarrow (x_0,y_0)}(cf (x,y))=cL$

3. $\lim_{(x,y)\rightarrow (x_0,y_0)}(f(x,y) g(x,y))=LM$

4. $\lim_{(x,y)\rightarrow (x_0,y_0)}\dfrac{f(x,y)}{g(x,y)}=\dfrac{L}{M} \quad \text{provided } M\neq 0$.

5. If $h(x)$ is a continuous function at $x=L$, then $\lim_{(x,y)\rightarrow (x_0,y_0)}h(f(x,y))=h(L)$.

6. $\lim_{(x,y)\rightarrow (x_0,y_0)}\left|f(x,y)\right|=|L|$.
• If the limit of a function exists, then it is unique.

In fact, this property is obvious, because $\lim_{(x,y)\to(x_0,y_0)} f(x,y)=L$ means that for all points $(x,y)$ near (but not equal to) $(x_0,y_0)$, the value of $f(x,y)$ is very close to $L$. So it is impossible for the value of $f(x,y)$ to be very close to two different numbers $L$ and $L’$ when $(x,y)$ is sufficiently close to $(x_0,y_0)$.

• Remember that when the limit of a function of a single variable at $x_0$ exists, the function should approach the same value as $x$ approaches $x_0$ from either left or right.
$$\lim_{x\rightarrow x_0} f(x)=L \Rightarrow \lim_{x\rightarrow x_0^+}f(x)=\lim_{x\rightarrow x_0^-}f(x)=L$$
For functions of several variables, if $\lim_{(x,y)\rightarrow (x_0,y_0)}f(x,y)=L$, then $f(x,y)$ should approach the same value no matter how $(x,y)$ approaches $(x_0,y_0)$ in the domain of $f$. Therefore:

If $f(x,y)$ approaches $L_1$ as $(x,y)$ approaches $(x_0,y_0)$ along a path, and if $f(x,y)$ approaches $L_2$ as $(x,y)$ approaches $(x_0,y_0)$ along a different path, and $L_1\neq L_2$, then $\lim_{(x,y)\rightarrow (x_0,y_0)}f(x,y)$ does not exist.

A special case of two different paths is when $(x,y)$ approaches $(x_0,y_0)$ along the the vertical line $x=x_0$ or the horizontal line $y=y_0$. In this case, we have two single variable limits:
$$\lim_{y\rightarrow y_0}f(x_0,y),$$
and
$$\lim_{x\rightarrow x_0}f(x,y_0).$$

Example 4

Does the following limit exist?
$$\lim_{(x,y)\rightarrow (0,0)}\frac{x+y}{x-y}$$

Solution
Because
$$\lim_{x\rightarrow 0} f(x,0)=\lim_{x\rightarrow 0}\left(\frac{x}{x}\right)=1$$
$$\lim_{y\rightarrow 0}f(0,y)=\lim_{y\rightarrow 0}\left(\frac{y}{-y}\right)=-1$$
are not equal, the limit does not exist. Graph of this function is shown in the following figure.

Example 5
Investigate
$$\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{x^2+y^2}.$$

Solution
At first, because
$$\lim_{x\rightarrow 0} f(x,0)=\lim_{x\rightarrow 0}0=0$$
$$\lim_{y\rightarrow 0}f(0,y)=\lim_{y\rightarrow 0}0=0$$
we might think that the limit exists and is equal to zero. But if we let $(x,y)$ approach $(0,0)$ along $y=mx$, we will realize
$$\lim_{x\rightarrow 0}\frac{x (mx)}{x^2+m^2x^2}=\lim_{x\rightarrow 0}\frac{mx^2}{(1+m^2)x^2}=\frac{m}{1+m^2}$$
depends on $m$. Therefore $\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{x^2+y^2}$ does not exist. The graph of this function is shown in the following figure.

Example 6
Investigate  $$\lim_{(x,y)\rightarrow (0,0)}\frac{xy^2}{x^2+y^4}.$$

Solution
If $(x,y)$ approaches $(0,0)$ along the lines $y=mx$, we will obtain
$\lim_{x\rightarrow 0}\frac{mx^3}{(1+m^2)x^4}=0.$ Can we conclude the limit exists and is equal to zero? Let's take another path to the origin: $y=x^2$. Along this path to the origin, we have
$$\lim_{x\rightarrow 0}\frac{x^4}{x^4}=1.$$
Because $\frac{xy^2}{x^2+y^4}$ approaches two different values $0$ and $1$ along two different paths to the origin, the limit does not exist.

Similar to the limit of functions of a single variable, we can use the sandwich theorem. According to this theorem (also called the squeeze theorem), if a function $f$ is sandwiched between two functions $g$ and $h$ and functions $g$ and $h$ approach to the same limit $L$ as $(x,y)\to(x_0,y_0)$, then $f$ also approaches to $L$ as $(x,y)\to(x_0,y_0)$. More precisely:

Theorem2  (Sandwich Theorem): If $g(x,y)\leq f(x,y)\leq h(x,y)$ for all points $(x,y)$ close to $(x_0,y_0)$ and
$$\lim_{(x,y)\rightarrow (x_0,y_0)}g(x,y)=\lim_{(x,y)\rightarrow (x_0,y_0)}h(x,y)=L$$
then
$$\lim_{(x,y)\rightarrow (x_0,y_0)}f(x,y)=L$$

This theorem is graphically shown in Figure 6.

Example 7

Consider the function
$$f(x,y)=\frac{x^5}{x^4+y^4}.$$
Find $\lim_{(x,y)\rightarrow (0,0)} f(x,y)$ if it exists.

Solution
We note that
$$\left|\frac{x^5}{x^4+y^4}\right|=|x|\left|\frac{x^4}{x^4+y^4}\right|\leq |x|.$$
Because $\lim_{(x,y)\rightarrow (0,0)} x=\lim_{(x,y)\rightarrow (0,0)} (-x)=0$, we conclude that
$\lim_{(x,y)\rightarrow (0,0)} f(x,y)=0$.

The graph of $f$ is shown in the following figure.

Example 8
Consider the function
$$g(x,y)=x\sin\left(\frac{1}{x^2+y}\right).$$
Find $\lim_{(x,y)\rightarrow (0,0)} g(x,y)$ if it exists.

Solution
Again because:
$$\left|x\sin\left(\frac{1}{x^2+y}\right)\right|\leq |x|$$
we conclude $\lim_{(x,y)\rightarrow (0,0)} g(x,y)=0$.

## Continuity

Remember that a function $g:U\subseteq\mathbb{R}\rightarrow\mathbb{R}$ is continuous at $x=a$ if $\lim_{x\rightarrow a}g(x)=g(a)$. Similarly, we define the continuity of multivariable functions.

Definition 2: A function $f$ is said to be  continuous at $(x_0,y_0)$ if

1. $f$ is defined at $(x_0,y_0)$.*
2. $\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=f(x_0,y_0)$

We say $f$ is continuous on a set $U$ if it is continuous at all points of $U$.

* In other words, $(x_0,y_0)$ is in the domain of $f$.

In Example 2, we saw that $f(x,y)=\frac{x^2y^2}{x^2+y^2}$ is not defined at $(0,0)$. Therefore $f$ is discontinuous at $(0,0)$. However, we can remove this discontinuity if we define $f(0,0)$ equal to its limit there, i.e.:
$$f(x,y)=\begin{cases} \dfrac{x^2y^2}{x^2+y^2} & \text{if } x\neq 0\\ \\ 0 & \text{if } x=0 \end{cases}$$
Similarly, we can remove the discontinuity of the function given in Example 3.

In general, if $f$ is not defined at $(x_0,y_0)$, but $\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)$ exists, we can remove the discontinuity by defining $f$ at this point as being equal to $\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)$. In this case, we say $f$ has a removable discontinuity at $(x_0,y_0)$.

Theorem 3: If $f(x,y)$ and $g(x,y)$ are two continuous functions at $(x_0,y_0)$ and $c$ is a scalar, then the following functions are continuous:

1.  $f(x,y)\pm g(x,y)$
2. $c f(x,y)$
3. $f(x,y) g(x,y)$
4. $f(x,y)/g(x,y)$               provided $g(x_0,y_0)\neq 0$

If $h(x)$ is continuous at $f(x_0,y_0)$, then $h\circ f (x,y)=h(f(x,y))$ is continuous at $(x_0,y_0)$; that is,
$$\lim_{(x,y)\rightarrow (x_0,y_0)} h(f(x,y))=h(f(x_0,y_0)).$$

## Limits and Continuity in n-space

To gain a better intuition, we focused on the concept of a limit on functions of two variables. However, what we studied in this section can be extended to functions of three or more variables if we consider the meaning of distance in $\mathbb{R}^n$ (for $n=1,2,3,\cdots$). If $\mathbf{a}=(a_1,\cdots,a_n)$ and $\mathbf{b}=(b_1,\cdots,b_n)$ are two points in $\mathbb{R}^n$, the distance between them is given by:
$$|\mathbf{a}-\mathbf{b}|=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}.$$
Therefore, for $f:U\subseteq\mathbb{R}^3\to\mathbb{R}$, the notation $\lim_{(x,y,z)\to(x_0,y_0,z_0)} f(x,y,z)=L$ means (a) every neighborhood of $(x_0,y_0,z_0)$ has points in the domain of $f$ other than $(x_0,y_0,z_0)$, and (b) for every $\epsilon>0$, there exists $\delta>0$ such that
$$(x,y,z)\in U \quad \text{and}\quad 0<|(x,y,z)-(x_0,y_0,z_0)|=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}<\delta$$
$$\Rightarrow |f(x,y,z)-L|<\epsilon.$$
Also, we say $f$ is continuous at $(x_0,y_0,z_0)$ if $\lim_{(x,y,z)\to(x_0,y_0,z_0)} f(x,y,z)=f(x_0,y_0,z_0)$.

Definition 3: Let $f:U\rightarrow \mathbb{R}$, where $U\subseteq \mathbb{R}^n$. We say $f(\mathbf{x})$ approaches $L$ as $\mathbf{x}=(x_1,\cdots,x_n)$ approaches $\mathbf{x}_0=(a_1,\cdots,a_n)$ and write
$$\lim_{\mathbf{x}\rightarrow\mathbf{x}_0} f(\mathbf{x})=L \quad (\text{or } f(\mathbf{x})\rightarrow L, \text{as } \mathbf{x}\rightarrow \mathbf{x}_0)$$if

1. every neighborhood of $\mathbf{x}_0$ contains at least one member of $U$ other than $\mathbf{x}_0$,
2. given any number $\epsilon>0$, there exists a number $\delta=\delta(\epsilon)>0$ such that $|f(\mathbf{x})-L|<\epsilon$ holds for any $\mathbf{x}$ in $U$ which satisfies $0<|\mathbf{x}-\mathbf{x}_0|<\epsilon$.

$f$ is called continuous at $\mathbf{x}_0$ if

$$\lim_{\mathbf{x}\rightarrow\mathbf{x}_0} f(\mathbf{x})=f(\mathbf{x}_0).$$

Again, the limit of a function at a point is unique; if along two paths to $\mathbf{x}_0$ the function approaches two different values, then its limit at $\mathbf{x}_0$ does not exist. The previous theorems in this section hold true when we switch from functions of two variables to functions of multiple variables. To determine the limit of functions of multiple variables, techniques similar to those for functions of two variables may be helpful. For example:

Example 9
Consider the function
$$f(x,y,z)=\frac{xyz}{x^2+y^2+z^2}$$
Find $\lim_{(x,y,z)\rightarrow (0,0,0)} f(x,y,z)$ if it exists.
Solution
Let’s use spherical coordinates (see Figure 8). Remember that in spherical coordinates:
$$x=\rho \cos\theta \sin \phi,\quad y=\rho \sin\theta \sin\phi,\quad z=\rho \cos\phi$$

So we can rewrite $f(x,y,z)$ in spherical coordinates as
$$f(x,y,z)=\frac{xyz}{x^2+y^2+z^2}=\frac{\rho^3 \sin^2\phi\cos\phi \cos\theta \sin\theta}{\rho^2}=\rho(\sin^2\phi\cos\phi \cos\theta \sin\theta)$$
Because the sine and cosine function always vary between $-1$ and $1$, we have
$$-\rho\leq\rho(\sin^2\phi\cos\phi \cos\theta \sin\theta)\leq \rho$$
and because $\lim_{(x,y,z)\rightarrow (0,0,0)}\rho=\lim_{\rho\rightarrow 0} \rho=0$, it follows from the squeeze theorem that $\lim_{(x,y,z)\rightarrow (0,0,0)} f(x,y,z)=0$