Section 14.2
Limits and Continuity of Multivariate Functions
We would like to be able to do calculus on multivariate functions; so we begin, as we did with
single variable functions, by defining limits.
However, there is some difficulty here. In the case of a function of one variable, ” lim f (x)”
x→x0
means that we investigate the way that the function behaves as one variable, x, changes. We can
allow x to approach the value x0 from at most two directions:
We would like a definition for
lim
f (x, y), but this presents some difficulty; we can allow
(x,y)→(x0 ,y0 )
the point (x, y) to approach (x0 , y0 ) from any of infinitely many directions. For instance, consider
lim
(x,y)→(1,1)
−x2 + y,
where the surface and the point (1, 1, 0) are graphed below:
If we approach (1, 1) along the line y = 0 (the x axis), the surface’s height appears to approach
z = 0:
1
Section 14.2
Approaching (1, 1) along the line y = x appears to give us the same result–the surface’s height
approaches z = 0:
We could even approach (1, 1) along y = x2 ; again, the function’s value appears to be approaching z = 0:
Although there are infinitely many ways to approach (0, 0), it appears that regardless of the
path we choose, the function’s value will be approaching z = 0. In this case, we would write
lim
(x,y)→(1,1)
−x2 + y = 0.
Indeed, in order for the concept of limit to make sense, we require that the value for the limit be
2
Section 14.2
the same regardless of the direction from which we approach (x0 , y0 ). In the graph below, consider
lim f (x, y) :
(x,y)→(0,0)
Approaching (0, 0) along, say, the x axis, the surface appears to be approaching z = 1.
However, if we approach (0, 0) along the y axis, the surface appears to be approaching z = 0.
In a case such as this, we say that the limit does not exist.
Although there is a formal definition for limits of multivariate functions in our book, we will
use the following informal definition:
3
Section 14.2
Definition 0.0.1. The function f (x, y) has limit z = L as (x, y) approaches (a, b), denoted
lim
f (x, y) = L,
(x,y)→(a,b)
if we can make f (x, y) as close to L as we like for every point (x, y) in a disk centered at (a, b) by
choosing the radius of the disk to be sufficiently small.
Earlier, we claimed that
lim
(x,y)→(1,1)
−x2 + y = 0.
Let’s draw an disk of radius 2 about (1, 1, 0):
We see that points on the surface within the bounds of the disk are fairly close to a height of 0.
If we make the disk’s radius smaller, say r = 1, points on the surface within the disk are even
closer to having a height of 0:
4
Section 14.2
The closer we ”zoom in” around (1, 1, 0), the more we can control how close the heights of the
points in the disk are to 0.
Continuity
Our definition for continuity of a multivariate function is very similar to that for a single variable
function:
Definition 0.0.2. A function f (x, y) is continuous at the point (a, b) if the following conditions
are satisfied:
1. f (a, b) exists (i.e., (a, b) is in the domain of f (x, y))
2.
lim
f (x, y) exists
(x,y)→(a,b)
3.
lim
f (x, y) = f (a, b).
(x,y)→(a,b)
In essence, a multivariate function is continuous at a point (x0 , y0 ) in its domain if the function’s
limit (its expected behavior) matches the function’s value (its actual behavior).
We need a practical method for evaluating limits of multivariate functions; fortunately, the
substitution rule for functions of one variable applies to multivariate functions:
Theorem 0.0.3. Substitution Rule for Limits
If f (x, y) is a continuous function and (x0 , y0 ) is in the domain of f (x, y), then
lim
(x,y)→(x0 ,y0 )
f (x, y) = f (x0 , y0 ).
Example:
cos(xyπ)
.
(x,y)→(3,1) x − y
Since the function is continuous and (3, 1) is in its domain, the limit is
Find
lim
cos(3 · 1π)
1
=− .
3−1
2
Properties of Limits of Multivariate Functions
Theorem 0.0.4. If L, M , and k are real numbers so that
lim
f (x, y) = L
and
(x,y)→(a,b)
lim
(x,y)→(a,b)
then
5
g(x, y) = M,
Section 14.2
1.
lim
(x,y)→(a,b)
2.
lim
(x,y)→(a,b)
3.
f (x, y) ± g(x, y) = L ± M .
f (x, y) · g(x, y) = L · M .
f (x, y)
L
=
if M ̸= 0.
g(x,
y)
M
(x,y)→(a,b)
lim
4.
lim
kf (x, y) = kL.
(x,y)→(a,b)
5.
lim
(x,y)→(a,b)
(f (x, y))r/s = Lr/s if s ̸= 0 and r and s have no common factors.
Unfortunately, if the substitution rule for limits is inapplicable, we have few options available
for determining the function’s limit. One of the few possibilities is to rewrite the function via
factoring or multiplying by conjugates.
Example
Determine
x−y−1
√
√ .
(x,y)→(2,1)
x−1− y
lim
Notice that (2, 1) is not in the domain of the function, since this point causes the function’s
denominator to be 0. However, if we multiply by the conjugate of the denominator, we can rewrite
the function so that we can actually determine the limit:
√
√
x−1+ y
x−y−1
x−y−1
√
√
√
=
·
√
√
√
x−1− y
x−1− y
x−1+ y
√
√
(x − y − 1)( x − 1 + y)
=
x−1−y
√
√
= x−1+ y
whenever y ̸= x − 1.
The function in the last line will have the same limit as our original function, so we evaluate
√
√
lim
x−1+ y =1+1
(x,y)→(2,1)
= 2.
So
lim
(x,y)→(2,1)
√
x−y−1
√ = 2.
x−1− y
If rewriting the function does not help in an attempt to evaluate the limit, we may alternatively
try to show that the limit does not exist; our method for doing this draws inspiration from the
issues we discussed at the beginning of the section. Specifically, we will attempt to show that we
get different ”limits” by approaching the point (x0 , y0 ) from different directions.
6
Section 14.2
Theorem 0.0.5. If the function f (x, y) has different limits along two different paths approaching
(a, b) (where both paths lie in the function’s domain), then
lim f (x, y) does not exist.
(x,y)→(a,b)
Looking at the limit along a specific path y = g(x) simplifies the problem by turning the multivariate function f (x, y) into a single variable function f (x, g(x)) (whose limit we can determine);
f (x, y) = f (x, g(x)) along the path g(x). If two different paths yield different ”limits”, then the
limit for f (x, y) does not exist.
In practice, the easiest way to use the theorem is to attempt to find two lines in the function’s
domain so that the function has different limits along the lines; however, even if the limit does not
exist, this approach is not always possible. As we will see in the second example below, we may
have to choose a more complicated path to find a different limit.
Example
Show that
x2 y 2
does not exist.
(x,y)→(0,0) x4 + 3y 4
lim
Let’s try evaluating the limit along the line y = x. Along this path, the function’s value is
x2 x2
x4
=
.
x4 + 3x4
4x4
So along this path, the limit is
x4
1
= lim
x→0 4x4
x→0 4
1
= .
4
lim
In other words, if we approach the point (0, 0) along the identity line, the function’s height
√
1
appears to approach . Let’s try doing the same thing for a different line, say y = 2x. Along
4
this path, the function’s value is
√
x2 ( 2x)2
2x4
2x4
√
= 4
=
.
x + 12x4
13x4
x4 + 3( 2x)4
So along this path, the limit is
2x4
2
= lim
x→0 13x4
x→0 13
2
= .
13
lim
√
Thus if we approach the point (0, 0) along the line y = 2x, the function’s height appears to
2
approach
. Since we’ve found two paths where the ”limits” are different, the function’s limit
13
does not exist.
7
Section 14.2
Find
x3 y
, if it exists.
(x,y)→(0,0) x6 + y 2
lim
It is clear that (0, 0) is not in the domain of the function, thus we can not use the substitution
rule to evaluate the limit. In addition, there is no good way to rewrite the function so that the
limit becomes easy to evaluate. These are really our only two options for determining that the
limit does exist, so let’s attempt to show that the limit does not exist.
We will again start by evaluating the limit along the line y = x. Along this path, the function’s
value is
x4
x3 x
=
.
x6 + x2
x6 + x2
So the limit along y = x is
x4
x2 (x2 )
=
lim
x→0 x6 + x2
x→0 x2 (x4 + 1)
x2
= lim 4
x→0 x + 1
0
=
1
= 0.
lim
So the limit along the identity line is 0. Let’s try doing the same thing for a different line, say
y = 2x. Along this path, the function’s value is
x3 (2x)
2x4
=
.
x6 + (2x)2
x6 + 4x2
The limit along this path is
2x4
x2 (2x2 )
=
lim
x→0 x6 + 4x2
x→0 x2 (x4 + 4)
2x2
= lim 4
x→0 x + 4
0
=
4
= 0.
lim
Unfortunately, the limit along y = 2x is precisely the same as the limit along y = x; this does
not help us at all, since we need to find two different ”limits” to conclude that the limit of the
original function does not exist. In fact, in this case, the limit along any line passing through (0, 0)
will be 0.
We may still suspect that the limit does not exist, so let’s try evaluating along a different path;
y = x3 is in the function’s domain whenever y ̸= 0, it’s easy to work with, and it is fundamentally
different from a line, so it is a good candidate for evaluating a limit. Along y = x3 , the function’s
value is
x3 (x3 )
x6
x6
=
=
.
x6 + (x3 )2
x6 + x6
2x6
8
Section 14.2
The limit along y = x3 is
x6
1
= lim
6
x→0 2x
x→0 2
1
= .
2
lim
We have found two paths paths which yield different limits, so we conclude that
does not exist.
9
x3 y
(x,y)→(0,0) x6 + y 2
lim