(1)(2)
2
lim
=
=
( x, y)→(1,2) x 2 + y 2
12 + 22
5
since f ( x, y) = xy is properly defined at the point (1,2).
x 2+ y 2
The major difference between limits in one variable and limits in two or more variables
has to do with how a point is approached. In the single-variable case, the statement “x → a”
means that x gets closer to the value a from two possible directions along the real number
line (see Figure 2.1.2(a)). In two dimensions, however, ( x, y) can approach a point ( a, b) along an infinite number of paths (see Figure 2.1.2(b)).
y
x
x
( a, b)
x
0
a
x
0
(a) x → a in R
(b) ( x, y) → ( a, b) in R2
Figure 2.1.2
“Approaching” a point in different dimensions
68
CHAPTER 2. FUNCTIONS OF SEVERAL VARIABLES
Example 2.7.
x y
lim
does not exist
( x, y)→(0,0) x 2 + y 2
Note that we can not simply substitute ( x, y) = (0,0) into the function, since doing so gives an
indeterminate form 0/0. To show that the limit does not exist, we will show that the function
approaches different values as ( x, y) approaches (0, 0) along different paths in R2. To see this,
suppose that ( x, y) → (0,0) along the positive x-axis, so that y = 0 along that path. Then
x y
x 0
f ( x, y) =
=
= 0
x 2 + y 2
x 2 + 02
along that path (since x > 0 in the denominator). But if ( x, y) → (0,0) along the straight line
y = x through the origin, for x > 0, then we see that
x y
x 2
1
f ( x, y) =
=
=
x 2 + y 2
x 2 + x 2
2
which means that f ( x, y) approaches different values as ( x, y) → (0,0) along different paths.
Hence the limit does not exist.
Limits of real-valued multivariable functions obey the same algebraic rules as in the
single-variable case, as shown in the following theorem, which we state without proof.
Theorem 2.1. Suppose that
lim
f ( x, y) and
lim
g( x, y) both exist, and that k is
( x, y)→( a, b)
( x, y)→( a, b)
some scalar. Then:
(a)
lim
[ f ( x, y) ± g( x, y)] =
lim
f ( x, y) ±
lim
g( x, y)
( x, y)→( a, b)
( x, y)→( a, b)
( x, y)→( a, b)
(b)
lim
k f ( x, y) = k
lim
f ( x, y)
( x, y)→( a, b)
( x, y)→( a, b)
(c)
lim
[ f ( x, y) g( x, y)] =
lim
f ( x, y)
lim
g( x, y)
( x, y)→( a, b)
( x, y)→( a, b)
( x, y)→( a, b)
lim
f ( x, y)
f ( x, y)
( x, y)→( a, b)
(d)
lim
=
if
lim
g( x, y) = 0
( x, y)→( a, b) g( x, y)
lim
g( x, y)
( x, y)→( a, b)
( x, y)→( a, b)
(e) If | f ( x, y) − L| ≤ g( x, y) for all ( x, y) and if
lim
g( x, y) = 0, then
lim
f ( x, y) = L.
( x, y)→( a, b)
( x, y)→( a, b)
Note that in part (e), it suffices to have | f ( x, y)− L| ≤ g( x, y) for all ( x, y) “sufficiently close”
to ( a, b) (but excluding ( a, b) itself).
2.1 Functions of Two or Three Variables
69
Example 2.8. Show that
y 4
lim
= 0.
( x, y)→(0,0) x 2 + y 2
Since substituting ( x, y) = (0,0) into the function gives the indeterminate form 0/0, we need
an alternate method for evaluating this limit. We will use Theorem 2.1(e). First, notice that
y 4 =
y 2 4 and so 0 ≤ y 4 ≤
x 2 + y 2 4 for all ( x, y). But
x 2 + y 2 4 = ( x 2 + y 2)2. Thus, for
all ( x, y) = (0,0) we have
y 4
( x 2 + y 2)2
≤
= x 2 + y 2 → 0 as ( x, y) → (0,0).
x 2 + y 2
x 2 + y 2
y 4
Therefore
lim
= 0.
( x, y)→(0,0) x 2 + y 2
Continuity can be defined similarly as in the single-variable case.
Definition 2.2. A real-valued function f ( x, y) with domain D in R2 is continuous at the
point ( a, b) in D if
lim
f ( x, y) = f ( a, b). We say that f ( x, y) is a continuous function if
( x, y)→( a, b)
it is continuous at every point in its domain D.
Unless indicated otherwise, you can assume that all the functions we deal with are con-
tinuous. In fact, we can modify the function from Example 2.8 so that it is continuous on all
of R2.
Example 2.9. Define a function f ( x, y) on all of R2 as follows:
0
if ( x, y)
= (0,0)
f ( x, y) =
y 4
if ( x, y) = (0,0)
x 2 + y 2
Then f ( x, y) is well-defined for all ( x, y) in R2 (i.e. there are no indeterminate forms for any
( x, y)), and we see that
b 4
lim
f ( x, y) =
= f ( a, b) for ( a, b) = (0,0).
( x, y)→( a, b)
a 2 + b 2
So since
lim
f ( x, y) = 0 = f (0,0) by Example 2.8,
( x, y)→(0,0)
then f ( x, y) is continuous on all of R2.
70
CHAPTER 2. FUNCTIONS OF SEVERAL VARIABLES
Exercises
A
For Exercises 1-6, state the domain and range of the given function.
1
1. f ( x, y) = x 2 + y 2 − 1
2. f ( x, y) = x 2 + y 2
x 2 + 1
3. f ( x, y) =
x 2 + y 2 − 4
4. f ( x, y) =
y
5. f ( x, y, z) = sin( xyz)
6. f ( x, y, z) =
( x − 1)( yz − 1)
For Exercises 7-18, evaluate the given limit.
7.
lim
cos( x y)
8.
lim
exy
( x, y)→(0,0)
( x, y)→(0,0)
x 2 − y 2
x y 2
9.
lim
10.
lim
( x, y)→(0,0) x 2 + y 2
( x, y)→(0,0) x 2 + y 4
x 2 − 2 xy + y 2
x y 2
11.
lim
12.
lim
( x, y)→(1,−1)
x − y
( x, y)→(0,0) x 2 + y 2
x 2 − y 2
x 2 − 2 xy + y 2
13.
lim
14.
lim
( x, y)→(1,1)
x − y
( x, y)→(0,0)
x − y
y 4 sin( x y)
1
15.
lim
16.
lim
( x 2 + y 2)cos
( x, y)→(0,0)
x 2 + y 2
( x, y)→(0,0)
x y
x
17.
lim
1
18.
lim
cos
( x, y)→(0,0) y
( x, y)→(0,0)
x y
B
19. Show that f ( x, y) = 1 e−( x 2+ y 2)/2 σ 2, for σ > 0, is constant on the circle of radius r > 0
2 πσ 2
centered at the origin. This function is called a Gaussian blur, and is used as a filter in
image processing software to produce a “blurred” effect.
20. Suppose that f ( x, y) ≤ f ( y, x) for all ( x, y) in R2. Show that f ( x, y) = f ( y, x) for all ( x, y) in R2.
21. Use the substitution r =
x 2 + y 2 to show that
sin
x 2 + y 2
lim
= 1 .
( x, y)→(0,0)
x 2 + y 2
( Hint: You will need to use L’Hôpital’s Rule for single-variable limits. )
C
22. Prove Theorem 2.1(a) in the case of addition. ( Hint: Use Definition 2.1. )
23. Prove Theorem 2.1(b).
2.2 Partial Derivatives
71
2.2 Partial Derivatives
Now that we have an idea of what functions of several variables are, and what a limit of
such a function is, we can start to develop an idea of a derivative of a function of two or more
variables. We will start with the notion of a partial derivative.
Definition 2.3. Let f ( x, y) be a real-valued function with domain D in R2, and let ( a, b) be
a point in D. Then the partial derivative of f at ( a, b) with respect to x, denoted by
∂ f ( a, b), is defined as
∂x
∂ f
f ( a + h, b) − f ( a, b)
( a, b) = lim
(2.2)
∂x
h→0
h
∂ f
and the partial derivative of f at ( a, b) with respect to y, denoted by
( a, b), is defined
∂y
as
∂ f
f ( a, b + h) − f ( a, b)
( a, b) = lim
.
(2.3)
∂y
h→0
h
Note: The symbol ∂ is pronounced “del” .1
Recall that the derivative of a function f ( x) can be interpreted as the rate of change of
that function in the (positive) x direction. From the definitions above, we can see that the
partial derivative of a function f ( x, y) with respect to x or y is the rate of change of f ( x, y) in the (positive) x or y direction, respectively. What this means is that the partial derivative of
a function f ( x, y) with respect to x can be calculated by treating the y variable as a constant,
and then simply differentiating f ( x, y) as if it were a function of x alone, using the usual
rules from single-variable calculus. Likewise, the partial derivative of f ( x, y) with respect to
y is obtained by treating the x variable as a constant and then differentiating f ( x, y) as if it
were a function of y alone.
∂ f
∂ f
Example 2.10. Find
( x, y) and
( x, y) for the function f ( x, y) = x 2 y + y 3.
∂x
∂y
Solution: Treating y as a constant and differentiating f ( x, y) with respect to x gives
∂ f ( x, y) =2 xy
∂x
and treating x as a constant and differentiating f ( x, y) with respect to y gives
∂ f ( x, y) = x 2+3 y 2 .
∂y
1It is not a Greek letter. The symbol was first used by the mathematicians A. Clairaut and L. Euler around
1740, to distinguish it from the letter d used for the “usual” derivative.
72
CHAPTER 2. FUNCTIONS OF SEVERAL VARIABLES
∂ f
∂ f
∂ f
∂ f
We will often simply write
and
instead of
( x, y) and
( x, y).
∂x
∂y
∂x
∂y
∂ f
∂ f
sin( x y 2)
Example 2.11. Find
and
for the function f ( x,