Limits and Continuity

We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to take the limit of a function of more than one variable, and what it means for a function of more than one variable to be continuous at a point in its domain. It turns out these concepts have aspects that just don’t occur with functions of one variable.

Limit of a Function of Two Variables

Before we can adapt this definition to define a limit of a function of two variables, we first need to see how to extend the idea of an open interval in one variable to an open interval in two variables.

disk appears in the definition of the limit of a function of two variables. If

δ

in the definition of a limit of a function of one variable. In one dimension, we express this restriction as

Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are an extension of the formulas in the limit laws theorem in The Limit Laws.

The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions.

Finding the Limit of a Function of Two Variables

Find each of the following limits:

$lim_{(x, y) \to (2, −1)} (x^{2} - 2 x y + 3 y^{2} - 4 x + 3 y - 6)$
$lim_{(x, y) \to (2, −1)} \frac{2 x + 3 y}{4 x - 3 y}$

First use the sum and difference laws to separate the terms:

$\begin{matrix} lim_{(x, y) \to (2, −1)} (x^{2} - 2 x y + 3 y^{2} - 4 x + 3 y - 6) \\ = (lim_{(x, y) \to (2, −1)} x^{2}) - (lim_{(x, y) \to (2, −1)} 2 x y) + (lim_{(x, y) \to (2, −1)} 3 y^{2}) - (lim_{(x, y) \to (2, −1)} 4 x) \\ + (lim_{(x, y) \to (2, −1)} 3 y) - (lim_{(x, y) \to (2, −1)} 6) . \end{matrix}$

Next, use the constant multiple law on the second, third, fourth, and fifth limits:

$\begin{matrix} = (lim_{(x, y) \to (2, −1)} x^{2}) - 2 (lim_{(x, y) \to (2, −1)} x y) + 3 (lim_{(x, y) \to (2, −1)} y^{2}) - 4 (lim_{(x, y) \to (2, −1)} x) \\ + 3 (lim_{(x, y) \to (2, −1)} y) - lim_{(x, y) \to (2, −1)} 6 . \end{matrix}$

Now, use the power law on the first and third limits, and the product law on the second limit:

$\begin{matrix} = {(lim_{(x, y) \to (2, −1)} x)}^{2} - 2 (lim_{(x, y) \to (2, −1)} x) (lim_{(x, y) \to (2, −1)} y) + 3 {(lim_{(x, y) \to (2, −1)} y)}^{2} \\ - 4 (lim_{(x, y) \to (2, −1)} x) + 3 (lim_{(x, y) \to (2, −1)} y) - lim_{(x, y) \to (2, −1)} 6 . \end{matrix}$

Last, use the identity laws on the first six limits and the constant law on the last limit:

$\begin{matrix} lim_{(x, y) \to (2, −1)} (x^{2} - 2 x y + 3 y^{2} - 4 x + 3 y - 6) & = {(2)}^{2} - 2 (2) (−1) + 3 {(−1)}^{2} - 4 (2) + 3 (−1) - 6 \\ = −6. \end{matrix}$
Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,

$\begin{matrix} lim_{(x, y) \to (2, −1)} (4 x - 3 y) & = lim_{(x, y) \to (2, −1)} 4 x - lim_{(x, y) \to (2, −1)} 3 y \\ = 4 (lim_{(x, y) \to (2, −1)} x) - 3 (lim_{(x, y) \to (2, −1)} y) \\ = 4 (2) - 3 (−1) = 11. \end{matrix}$

Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:

$\begin{matrix} lim_{(x, y) \to (2, −1)} (2 x + 3 y) & = lim_{(x, y) \to (2, −1)} 2 x + lim_{(x, y) \to (2, −1)} 3 y \\ = 2 (lim_{(x, y) \to (2, −1)} x) + 3 (lim_{(x, y) \to (2, −1)} y) \\ = 2 (2) + 3 (−1) \\ = 1. \end{matrix}$

Therefore, according to the quotient law we have

$lim_{(x, y) \to (2, −1)} \frac{2 x + 3 y}{4 x - 3 y} = \frac{lim_{(x, y) \to (2, −1)} (2 x + 3 y)}{lim_{(x, y) \to (2, −1)} (4 x - 3 y)} = \frac{1}{11} .$

Since we are taking the limit of a function of two variables, the point

(a, b)

and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward

(a, b) .

If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken.

Limits That Fail to Exist

Show that neither of the following limits exist:

$lim_{(x, y) \to (0, 0)} \frac{2 x y}{3 x^{2} + y^{2}}$
$lim_{(x, y) \to (0, 0)} \frac{4 x y^{2}}{x^{2} + 3 y^{4}}$

The domain of the function $f (x, y) = \frac{2 x y}{3 x^{2} + y^{2}}$
consists of all points in the
$x y -plane$
except for the point
$(0, 0)$
([link]). To show that the limit does not exist as
$(x, y)$
approaches
$(0, 0),$
we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point
$(0, 0) .$
First, consider the line
$y = 0$
in the
$x y -plane.$
Substituting
$y = 0$
into
$f (x, y)$
gives

$f (x, 0) = \frac{2 x (0)}{3 x^{2} + 0^{2}} = 0$

for any value of
$x .$
Therefore the value of
$f$
remains constant for any point on the
$x -axis,$
and as
$y$
approaches zero, the function remains fixed at zero.

Next, consider the line
$y = x .$
Substituting
$y = x$
into
$f (x, y)$
gives

$f (x, x) = \frac{2 x (x)}{3 x^{2} + x^{2}} = \frac{2 x^{2}}{4 x^{2}} = \frac{1}{2} .$

This is true for any point on the line
$y = x .$
If we let
$x$
approach zero while staying on this line, the value of the function remains fixed at
$\frac{1}{2},$
regardless of how small
$x$
is.

Choose a value for
$ε$
that is less than
$1 / 2$
—say,
$1 / 4 .$
Then, no matter how small a
$δ$
disk we draw around
$(0, 0),$
the values of
$f (x, y)$
for points inside that
$δ$
disk will include both
$0$
and
$\frac{1}{2} .$
Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.

In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the
$x -axis$
(i.e.,
$y = 0),$
then the function remains fixed at zero. The same is true for the
$y -axis.$
Suppose we approach the origin along a straight line of slope
$k .$
The equation of this line is
$y = k x .$
Then the limit becomes

$\begin{matrix} lim_{(x, y) \to (0, 0)} \frac{4 x y^{2}}{x^{2} + 3 y^{4}} & = lim_{(x, y) \to (0, 0)} \frac{4 x {(k x)}^{2}}{x^{2} + 3 {(k x)}^{4}} \\ = lim_{(x, y) \to (0, 0)} \frac{4 k^{2} x^{3}}{x^{2} + 3 k^{4} x^{4}} \\ = lim_{(x, y) \to (0, 0)} \frac{4 k^{2} x}{1 + 3 k^{4} x^{2}} \\ = \frac{lim_{(x, y) \to (0, 0)} (4 k^{2} x)}{lim_{(x, y) \to (0, 0)} (1 + 3 k^{4} x^{2})} \\ = 0 \end{matrix}$

regardless of the value of
$k .$
It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation
$x = y^{2} .$
Substituting
$y^{2}$
in place of
$x$
in
$f (x, y)$
gives

$\begin{matrix} lim_{(x, y) \to (0, 0)} \frac{4 x y^{2}}{x^{2} + 3 y^{4}} & = lim_{(x, y) \to (0, 0)} \frac{4 (y^{2}) y^{2}}{{(y^{2})}^{2} + 3 y^{4}} \\ = lim_{(x, y) \to (0, 0)} \frac{4 y^{4}}{y^{4} + 3 y^{4}} \\ = lim_{(x, y) \to (0, 0)} 1 \\ = 1. \end{matrix}$

By the same logic in a., it is impossible to find a
$δ$
disk around the origin that satisfies the definition of the limit for any value of
$ε < 1 .$
Therefore,
$lim_{(x, y) \to (0, 0)} \frac{4 x y^{2}}{x^{2} + 3 y^{4}}$
does not exist.

Interior Points and Boundary Points

To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology.

disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a

δ

disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the

δ disk

is not contained inside the domain. By definition, some of the points of the

δ disk

are inside the domain and some are outside. Therefore, we need only consider points that are inside both the

δ

disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point.

Continuity of Functions of Two Variables

In Continuity, we defined the continuity of a function of one variable and saw how it relied on the limit of a function of one variable. In particular, three conditions are necessary for

f (x)

These three conditions are necessary for continuity of a function of two variables as well.

Continuity of a function of any number of variables can also be defined in terms of delta and epsilon. A function of two variables is continuous at a point

(x_{0}, y_{0})

This definition can be combined with the formal definition (that is, the epsilon–delta definition) of continuity of a function of one variable to prove the following theorems:

Let’s now use the previous theorems to show continuity of functions in the following examples.

Functions of Three or More Variables

The limit of a function of three or more variables occurs readily in applications. For example, suppose we have a function

f (x, y, z)

disk into more than two dimensions. Then, the ideas of the limit of a function of three or more variables and the continuity of a function of three or more variables are very similar to the definitions given earlier for a function of two variables.

To show that a limit of a function of three variables exists at a point

(x_{0}, y_{0}, z_{0}),

the value of the function at that point is arbitrarily close to a fixed value (the limit value). All the limit laws for functions of two variables hold for functions of more than two variables as well.

Key Concepts

For the following exercises, find the limit of the function.

lim_{(x, y) \to (1, 2)} x

lim_{(x, y) \to (1, 2)} \frac{5 x^{2} y}{x^{2} + y^{2}}

2.0

Show that the limit $lim_{(x, y) \to (0, 0)} \frac{5 x^{2} y}{x^{2} + y^{2}}$

exists and is the same along the paths: $y -axis$

and $x -axis,$

and along $y = x .$

For the following exercises, evaluate the limits at the indicated values of $x and y .$

If the limit does not exist, state this and explain why the limit does not exist.

lim_{(x, y) \to (0, 0)} \frac{4 x^{2} + 10 y^{2} + 4}{4 x^{2} - 10 y^{2} + 6}

\frac{2}{3}

lim_{(x, y) \to (11, 13)} \sqrt{\frac{1}{x y}}

lim_{(x, y) \to (0, 1)} \frac{y^{2} sin x}{x}

1

lim_{(x, y) \to (0, 0)} sin (\frac{x^{8} + y^{7}}{x - y + 10})

lim_{(x, y) \to (π / 4, 1)} \frac{y tan x}{y + 1}

\frac{1}{2}

lim_{(x, y) \to (0, π / 4)} \frac{sec x + 2}{3 x - tan y}

lim_{(x, y) \to (2, 5)} (\frac{1}{x} - \frac{5}{y})

- \frac{1}{2}

lim_{(x, y) \to (4, 4)} x ln y

lim_{(x, y) \to (4, 4)} e^{− x^{2} - y^{2}}

e^{−32}

lim_{(x, y) \to (0, 0)} \sqrt{9 - x^{2} - y^{2}}

lim_{(x, y) \to (1, 2)} (x^{2} y^{3} - x^{3} y^{2} + 3 x + 2 y)

11.0

lim_{(x, y) \to (π, π)} x sin (\frac{x + y}{4})

lim_{(x, y) \to (0, 0)} \frac{x y + 1}{x^{2} + y^{2} + 1}

1.0

lim_{(x, y) \to (0, 0)} \frac{x^{2} + y^{2}}{\sqrt{x^{2} + y^{2} + 1} - 1}

lim_{(x, y) \to (0, 0)} ln (x^{2} + y^{2})

The limit does not exist because when $x$

and $y$

both approach zero, the function approaches $ln 0,$

which is undefined (approaches negative infinity).

For the following exercises, complete the statement.

A point $(x_{0}, y_{0})$

in a plane region $R$

is an interior point of $R$

if _________________.

A point $(x_{0}, y_{0})$

in a plane region $R$

is called a boundary point of $R$

if ___________.

every open disk centered at $(x_{0}, y_{0})$

contains points inside $R$

and outside $R$

For the following exercises, use algebraic techniques to evaluate the limit.

lim_{(x, y) \to (2, 1)} \frac{x - y - 1}{\sqrt{x - y} - 1}

lim_{(x, y) \to (0, 0)} \frac{x^{4} - 4 y^{4}}{x^{2} + 2 y^{2}}

0.0

lim_{(x, y) \to (0, 0)} \frac{x^{3} - y^{3}}{x - y}

lim_{(x, y) \to (0, 0)} \frac{x^{2} - x y}{\sqrt{x} - \sqrt{y}}

0.00

For the following exercises, evaluate the limits of the functions of three variables.

lim_{(x, y, z) \to (1, 2, 3)} \frac{x z^{2} - y^{2} z}{x y z - 1}

lim_{(x, y, z) \to (0, 0, 0)} \frac{x^{2} - y^{2} - z^{2}}{x^{2} + y^{2} - z^{2}}

The limit does not exist.

For the following exercises, evaluate the limit of the function by determining the value the function approaches along the indicated paths. If the limit does not exist, explain why not.

lim_{(x, y) \to (0, 0)} \frac{x y + y^{3}}{x^{2} + y^{2}}

Along the $x -axis$ $(y = 0)$
Along the $y -axis$ $(x = 0)$
Along the path $y = 2 x$

Evaluate $lim_{(x, y) \to (0, 0)} \frac{x y + y^{3}}{x^{2} + y^{2}}$

using the results of previous problem.

The limit does not exist. The function approaches two different values along different paths.

lim_{(x, y) \to (0, 0)} \frac{x^{2} y}{x^{4} + y^{2}}

Along the x-axis $(y = 0)$
Along the y-axis $(x = 0)$
Along the path $y = x^{2}$

Evaluate $lim_{(x, y) \to (0, 0)} \frac{x^{2} y}{x^{4} + y^{2}}$

using the results of previous problem.

The limit does not exist because the function approaches two different values along the paths.

Discuss the continuity of the following functions. Find the largest region in the $x y -plane$

in which the following functions are continuous.

f (x, y) = sin (x y)

f (x, y) = ln (x + y)

The function $f$

is continuous in the region $y > − x .$

f (x, y) = e^{3 x y}

f (x, y) = \frac{1}{x y}

The function $f$

is continuous at all points in the $x y -plane$

except at $(0, 0) .$

For the following exercises, determine the region in which the function is continuous. Explain your answer.

f (x, y) = \frac{x^{2} y}{x^{2} + y^{2}}

f (x, y) = {\begin{matrix} \frac{x^{2} y}{x^{2} + y^{2}} & if (x, y) \neq (0, 0) \\ 0 & if (x, y) = (0, 0) \end{matrix}}

(Hint: Show that the function approaches different values along two different paths.)

The function is continuous at $(0, 0)$

since the limit of the function at $(0, 0)$

is $0,$

the same value of $f (0, 0) .$

f (x, y) = \frac{sin (x^{2} + y^{2})}{x^{2} + y^{2}}

Determine whether $g (x, y) = \frac{x^{2} - y^{2}}{x^{2} + y^{2}}$

is continuous at $(0, 0) .$

The function is discontinuous at $(0, 0) .$

The limit at $(0, 0)$

fails to exist and $g (0, 0)$

does not exist.

Create a plot using graphing software to determine where the limit does not exist. Determine the region of the coordinate plane in which $f (x, y) = \frac{1}{x^{2} - y}$

is continuous.

Determine the region of the $x y -plane$

in which the composite function $g (x, y) = arctan (\frac{x y^{2}}{x + y})$

is continuous. Use technology to support your conclusion.

Since the function $arctan x$

is continuous over $(− \infty, \infty),$

g (x, y) = arctan (\frac{x y^{2}}{x + y})

is continuous where $z = \frac{x y^{2}}{x + y}$

is continuous. The inner function $z$

is continuous on all points of the $x y -plane$

except where $y = − x .$

Thus, $g (x, y) = arctan (\frac{x y^{2}}{x + y})$

is continuous on all points of the coordinate plane except at points at which $y = − x .$

Determine the region of the $x y -plane$

in which $f (x, y) = ln (x^{2} + y^{2} - 1)$

is continuous. Use technology to support your conclusion. (Hint: Choose the range of values for $x and y$

carefully!)

At what points in space is $g (x, y, z) = x^{2} + y^{2} - 2 z^{2}$

continuous?

All points $P (x, y, z)$

in space

At what points in space is $g (x, y, z) = \frac{1}{x^{2} + z^{2} - 1}$

continuous?

Show that $lim_{(x, y) \to (0, 0)} \frac{1}{x^{2} + y^{2}}$

does not exist at $(0, 0)$

by plotting the graph of the function.

The graph increases without bound as $x and y$

both approach zero.* * *

On xyz space, there is a shape drawn that decreases to 0 as x and y increase or decrease but that increases greatly closer to the origin. It increases to such an extent that the graph is cut off above z = 10, which coincides with a circle of radius 0.6 around (0, 0, 10).

[T] Evaluate $lim_{(x, y) \to (0, 0)} \frac{− x y^{2}}{x^{2} + y^{4}}$

by plotting the function using a CAS. Determine analytically the limit along the path $x = y^{2} .$

[T]

Use a CAS to draw a contour map of $z = \sqrt{9 - x^{2} - y^{2}} .$
What is the name of the geometric shape of the level curves?
Give the general equation of the level curves.
What is the maximum value of $z ?$
What is the domain of the function?
What is the range of the function?

a.* * *

A series of five concentric circles, with radii 3, 2.75, 2.5, 2.2, and 1.75. The areas between the circles are colored, with the darkest color between the circles of radii 3 and 2.75.

b. The level curves are circles centered at $(0, 0)$

with radius $9 - c .$

c. $x^{2} + y^{2} = 9 - c$

d. $z = 3$

e. ${(x, y) \in ℝ^{2} \| x^{2} + y^{2} \leq 9}$

f. ${z \| 0 \leq z \leq 3}$

True or False: If we evaluate $lim_{(x, y) \to (0, 0)} f (x)$

along several paths and each time the limit is $1,$

we can conclude that $lim_{(x, y) \to (0, 0)} f (x) = 1 .$

Use polar coordinates to find $lim_{(x, y) \to (0, 0)} \frac{sin \sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2}}} .$

You can also find the limit using L’Hôpital’s rule.

1.0

Use polar coordinates to find $lim_{(x, y) \to (0, 0)} cos (x^{2} + y^{2}) .$

Discuss the continuity of $f (g (x, y))$

where $f (t) = 1 / t$

and $g (x, y) = 2 x - 5 y .$

f (g (x, y))

is continuous at all points $(x, y)$

that are not on the line $2 x - 5 y = 0 .$

Given $f (x, y) = x^{2} - 4 y,$

find $lim_{h \to 0} \frac{f (x + h, y) - f (x, y)}{h} .$