OBJECTIVES

When you finish this section, you should be able to:

1. Define the limit of a function of several variables (p. 819)
2. Find a limit using properties of limits (p. 821)
3. Examine when limits exist (p. 822)
4. Determine whether a function is continuous (p. 824)

Let $f\,$ be a function of one variable defined everywhere in an open interval containing $c$, except possibly at $c$. Then the limit of $f(x)$ as $x$ approaches $c$ is $L$, written $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \lim\limits_{x\rightarrow c}f(x)=L }}$$

if for any number $\varepsilon >0$, there is a number $\delta > 0$, so that $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \hbox{whenever }\ 0<\left\vert x-c\right\vert <\delta \qquad \hbox{then}\qquad \vert f(x)-L\vert <\varepsilon }}$$

DEFINITION $\delta$-Neighborhood; Disk; Ball

Let $P_{0}$ be a point in the plane or in space. A $\delta$ -neighborhood of $P_{0}$ consists of all the points $P$ that lie within a distance $\delta$ of $P_{0}$. That is, a $\delta$-neighborhood consists of all points $P$ for which $d(P, P_{0})<\delta$, where $d( P,P_{0})$ is the distance from $P$ to $P_{0}.$

• In the plane, a $\delta$-neighborhood of $P_{0}$ is called a disk of radius $\delta$ centered at $P_{0}.$
• In space, a $\delta$-neighborhood of $P_{0}$ is called a ball of radius $\delta$ centered at $P_{0}$.

NOTE

On a real line, a $\delta$-neighborhood of a number $P_{0}$ is an open interval with $P_{0}$ as midpoint.

DEFINITION Limit of a Function of Two Variables

Let $f$ be a function of two variables defined at each point in some disk centered at $(x_{0},\,y_{0})$, except possibly at $(x_{0},y_{0})$. Then the limit of $f(x,y)$ as $(x,y)$ approaches $(x_{0},y_{0})$ is $L$ if, for any given number $\varepsilon >0$, there is a number $\delta >0$, so that $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \hbox{whenever }\ 0<\sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}}<\delta \qquad \hbox{then} \qquad \vert f(x,y) -L\vert <\varepsilon }}$$

Then we say, “The limit exists”, and write $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}f(x,y)=L }}$$

DEFINITION Limit of a Function of Three Variables

Let $f$ be a function of three variables defined at each point in some ball centered at $(x_{0},y_{0},z_{0})$, except possibly at $(x_{0},y_{0},z_{0})$. Then the limit of $f(x,y,z)$ as $(x,y,z)$ approaches $(x_{0},y_{0},z_{0})$ is $L$ if, for any given number $\varepsilon >0$, there is a number $\delta >0$, so that $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \hbox{whenever }\ 0<\sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}} <\delta \quad \hbox{then}\ \vert f(x,y,z) -L\vert <\varepsilon }}$$

Then we say, “The limit exists”, and write $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \lim\limits_{(x, y, z)\rightarrow (x_{0}, y_{0}, z_{0})}f(x,y,z)=L }}$$

THEOREM Algebraic Properties of Limits (Two Variables)

Let $f$ and $g$ be functions of two variables for which $$\lim_{(x, y)\rightarrow (x_{0}, y_{0})}f(x,y)=L\qquad \hbox{and}\qquad \lim_{(x, y)\rightarrow (x_{0}, y_{0})}g(x,y)=M$$

where $L$ and $M$ are two real numbers.

• Limit of a sum: $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{eqnarray*} \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}[f(x,y)+g(x,y)]&=&\lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}f(x,y) +\lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}g(x,y)\\ &=&L+M \end{eqnarray*}}}$$
• Limit of a difference: $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \begin{eqnarray*} \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}[f(x,y)-g(x,y)]&=&\lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}f(x,y) -\lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}g(x,y)\\ &=&L-M \end{eqnarray*}}}$$
• Limit of a constant times a function: If $k$ is any real number, then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{eqnarray*} \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}[k\,f(x,y)]=k\left[ \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}f(x,y)\right] =kL \end{eqnarray*}}}$$
• Limit of a product: $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{eqnarray*} \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}[f(x,y)g(x,y)]&=&\left[ \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}f(x,y)\right]\; \left[ \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}g(x,y)\right]\\[4pt] &=&LM \end{eqnarray*}}}$$
• Limit of a quotient: If $\lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}g(x,y)=M\neq 0$, $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}\dfrac{f(x,y)}{g(x,y) }=\dfrac{ \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}f(x,y)}{\lim\limits_{(x, y) \rightarrow (x_{0}, y_{0})}g(x,y)}=\dfrac{L}{M} }}$$
• Limit of a constant: If $c$ is a constant, $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}c=c }}$$
• Limit of a function of a single variable: If $f(x,y) =h(x) ,$ then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}f(x,y)=\lim\limits_{x\rightarrow x_{0}}h(x) }}$$
  If $f(x,y) =g(y) ,$ then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \lim\limits_{(x, y)\rightarrow (x_{0}, y_{0})}f(x,y)=\lim\limits_{y\rightarrow y_{0}}g(y) }}$$

Find each limit:

Solution (a) $$\begin{eqnarray*} \lim_{(x, y)\rightarrow (1, 2)}(3x^{2}+2xy+y^{2})& =&\lim_{(x, y)\rightarrow (1, 2)}( 3x^{2}) +\lim_{(x, y)\rightarrow (1, 2)}( 2xy) +\lim_{(x, y)\rightarrow (1, 2)}y^{2} \\[8pt] & =&3\lim_{x\rightarrow 1}x^{2}+\Big( \lim_{(x, y)\rightarrow (1, 2)}( 2x) \Big) \Big( \lim_{(x, y)\rightarrow (1, 2)}y\Big) +\lim_{y\rightarrow 2}y^{2} \\[8pt] &=& 3+2\Big( \lim_{x\rightarrow 1}x\Big) \Big( \lim_{y\rightarrow 2}y\Big) +4=3+2\cdot 1\cdot 2+4=11 \end{eqnarray*}$$

(b) Since $\lim\limits_{(x, y)\rightarrow (1, 2)} (x^2+y^2) = \lim\limits_{x\rightarrow 1} x^2 + \lim\limits_{y\rightarrow 2} y^2 =5\neq 0$, we use the limit of a quotient: $$\begin{eqnarray*} \lim_{(x, y)\rightarrow (1, 2)}\frac{xy}{x^{2}+y^{2}} &=&\frac{ \lim\limits_{(x, y)\rightarrow (1, 2)}(xy) }{\lim\limits_{(x, y) \rightarrow (1, 2)}(x^{2}+y^{2})} =\frac{\Big( \lim\limits_{x\rightarrow 1}x\Big) \Big( \lim\limits_{y\rightarrow 2}y\Big) }{\lim\limits_{x\rightarrow 1}x^{2}+\lim\limits_{y\rightarrow 2}y^{2}}=\frac{1\cdot 2}{1+4}=\frac{2}{5} \end{eqnarray*}$$

(c) Look at the denominator. Since $\lim\limits_{(x, y) \rightarrow (3,-1)}( x^{2}+9y^{2}) =18,$ we can use the limit of a quotient property. Then $$\begin{eqnarray*} \lim\limits_{(x, y)\rightarrow (3,-1)}\dfrac{x^{2}+2xy-3y^{2}}{x^{2}+9y^{2}} &=&\dfrac{\lim\limits_{(x, y)\rightarrow (3,-1)}(x^{2} + 2xy - 3y^{2}) }{\lim\limits_{(x, y)\rightarrow (3,-1)}(x^{2} + 9y^{2}) }\\[8pt] &=&\frac{\lim\limits_{x\rightarrow 3}x^{2} + 2\Big( \lim\limits_{x\rightarrow 3}x\Big) \Big( \lim\limits_{y\rightarrow -1}y\Big) - 3 \lim\limits_{y\rightarrow -1}y^{2}}{18}\\[8pt] &=&\dfrac{9 + 2(3)(-1) - 3(1)}{18}=0 \end{eqnarray*}$$

NOW WORK

Problems 9 and 21.

THEOREM

If $\lim\limits_{P\rightarrow P_{0}}f(P)$ is computed along two different curves containing $P_{0}$ and if two different answers are obtained, then $\lim\limits_{P\rightarrow P_{0}}f(P)$ does not exist.

Show that $\lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{xy}{ x^{2}+y^{2}}$ does not exist.

Solution  Look at the denominator. Since $\lim\limits_{(x, y) \rightarrow (0, 0)}( x^{2}+y^{2}) =0$, we cannot use the limit of a quotient to investigate the limit. We investigate the limit using two curves that contain $(0,0)$, say, $y=2x$ and $y=-x.$

First we investigate the limit using the line $y=2x$. Then $$\begin{equation*} {\rm Using}\; y=2x: \quad \lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x\left( 2x\right) }{x^{2}+\left( 2x\right) ^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ 2x^{2}}{x^{2}+4x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{2x^{2}}{5x^{2}}= \dfrac{2}{5} \end{equation*}$$

Figure 23 $f(x,y) =\dfrac{xy}{x^{2}+y^{2}}$

Now we investigate the limit using the line $y=-x$. $$\begin{equation*} {\rm Using}\; y=-x: \quad \lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x\left( -x\right) }{x^{2}+\left( -x\right) ^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ -x^{2}}{x^{2}+x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{-x^{2}}{2x^{2}}=- \dfrac{1}{2} \end{equation*}$$

Since two different answers are obtained, the limit does not exist.

NOW WORK

Problem 25.

Show that $\lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x^{2}y}{x^{4}+y^{2}}$ does not exist.

Solution  We investigate the limit using curves that contain $(0,0).$ Based on our experience with the function in Example 2, we can simultaneously test all nonvertical lines that contain $(0,0)$ by using $y=mx$. Then $$\begin{equation*} {\rm Using}\; y=mx: \quad \lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x^{2}\left( mx\right) }{x^{4}+\left( mx\right) ^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ mx^{3}}{x^{4}+m^{2}x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{mx}{x^{2}+m^{2}} =0 \end{equation*}$$

Along every nonvertical line containing the point $(0,0)$, the limit is $0$. But this does not mean that $\lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{ x^{2}y}{x^{4}+y^{2}}$ is $0$. Suppose we use the curve $y=x^{2}$. Then $$\begin{equation*} {\rm Using}\; y=x^{2}: \quad \lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x^{4}}{ x^{4}+x^{4}}=\lim\limits_{x\rightarrow 0}\dfrac{x^{4}}{2x^{4}}=\dfrac{1 }{2} \end{equation*}$$

Since two different curves that contain $(0,0)$ result in different values for the limit, the limit does not exist.

NOW WORK

Problem 33.

Use the $\varepsilon$-$\delta$ definition of a limit to show that $\lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{2xy^{2}}{x^{2}+y^{2}}=0.$

Solution  Given a number $\varepsilon >0$, we seek a number $\delta >0$, so that $$\begin{equation*} \hbox{whenever}\quad 0<\sqrt{(x-0) ^{2}+( y-0) ^{2}} <\delta \qquad \hbox{then}\quad \left\vert \dfrac{2xy^{2}}{x^{2}+y^{2}} -0\right\vert <\varepsilon \end{equation*}$$

That is, $$\begin{equation*} \hbox{whenever }\quad 0<\sqrt{x^{2}+y^{2}}<\delta \quad \hbox{then }\quad \left\vert \dfrac{2xy^{2}}{x^{2}+y^{2}}\right\vert <\varepsilon \end{equation*}$$

We need to find a connection between $\dfrac{2xy^{2}}{x^{2}+y^{2}}$ and $\sqrt{x^{2}+y^{2}}.$ We begin with the observation that since $x^{2}\geq 0,$ $$\begin{equation*} y^{2}\leq x^{2}+y^{2}\hbox{ } \end{equation*}$$

So, for all points $(x,y)$ not equal to $(0,0),$ we have $$\dfrac{y^{2}}{x^{2}+y^{2}}\leq 1\qquad {\color{#0066A7}{\hbox{Divide both sides by \(x^2+y^2\).}}}$$

Now $$\begin{equation*} \left\vert \frac{2xy^{2}}{x^{2}+y^{2}}\right\vert =\frac{2\vert x\vert y^{2}}{x^{2}+y^{2}}=2\vert x\vert \cdot \dfrac{ y^{2}}{x^{2}+y^{2}}\leq 2\left\vert x\right\vert \cdot 1=2\sqrt{x^{2}}\leq 2 \sqrt{x^{2}+y^{2}} \end{equation*}$$

Given $\varepsilon >0$, we choose $\delta \leq \dfrac{\varepsilon }{2}$. Then whenever $0<\sqrt{x^{2}+y^{2}}<\delta$, we have $$\begin{equation*} \left\vert \frac{2xy^{2}}{x^{2}+y^{2}}\right\vert \leq 2\sqrt{x^{2}+y^{2}} <2\delta \leq \varepsilon \end{equation*}$$

That is, $$\lim_{(x, y)\rightarrow (0, 0)}\left\vert \frac{2xy^{2}}{x^{2}+y^{2}} \right\vert =0 %%\tag{$\blacksquare \(}$$

NOW WORK

Problem 63.

NEED TO REVIEW?

Continuity for functions of one variable is discussed in Section 1.3, pp. 93-100.

DEFINITION Interior Point; Boundary Point; Open Set; Closed Set

Let $S$ be a set of points in the plane or in space.

• A point $P_{0}$ is called an interior point of $S$ if there is a $\delta$-neighborhood of $P_{0}$ that lies entirely in $S$.
• A point $P_{0}$ is called a boundary point of $S$ if every $\delta$-neighborhood of $P_{0}$ contains points both in $S$ and not in $S$. A boundary point of $S$ may be in $S$ or not in $S.$
• A set $S$ is called an open set if every point of $S$ is an interior point.
• A set $S$ is called a closed set if it contains all its boundary points.
• A set $S$ that contains some, but not all, of its boundary points is neither open nor closed.

NOTE

You may find it helpful to compare the definition given here with the one on p. 93 in Chapter 1.

DEFINITION Continuity at a Point $P_{0}$

Let $f$ be a function whose domain $D$ is some subset of points in the plane, and let $P_{0}$ be an interior point of $D.$ Then $f$ is continuous at $P_{0}$ if the following two conditions are met:

• $\lim\limits_{P\rightarrow P_{0}}f(P)$ exists
• $\lim\limits_{P\rightarrow P_{0}}f(P)=f(P_{0})$

DEFINITION Continuity on a Domain

A function $f$ is continuous on its domain if it is continuous at every point in its domain. Such functions are referred to as continuous functions.

• The sum, difference, product, or composition of two continuous functions is continuous.
• The quotient of two functions that are continuous at a point $P_{0}$ is also continuous at $P_{0}$, provided the denominator is not $0$ at $P_{0}.$

Figure 28

Determine where the following functions are continuous:

Solution (a) Since $f$ is a polynomial function, it is continuous on its domain, every point $(x,y)$ in the plane, as shown in Figure 28(a).

(b) $R$ is a rational function and is continuous on its domain, all points $(x,y)$ in the plane, except those for which $xy=1.$ The graph of $R$ will have no points on the cylinder $xy=1$, as shown in Figure 28(b).

NOW WORK

Problems 37 and 39.

• The composite of a continuous function of one variable $z=f( t)$ and a continuous function of two variables $t=g(x,y)$ is continuous on its domain.

NOW WORK

Problem 43.

• If $z=f(x,y)$ is continuous at the point $(x_{0},y_{0}) ,$ then $\lim\limits_{(x,y) \rightarrow (x_{0},y_{0}) }f(x,y) =f(x_{0},y_{0})$

NOW WORK

Problem 53.