Continuous Functions of Several Variables
Limits of Functions of Several Variables
Math 301
Lecture 2
Winter 2020
Rami Younes
Limits of Functions of Several Variables
Continuous Functions of Several Variables
Lecture Outline
1
Continuous Functions of Several Variables
Limits and Continuity
Limits Using Polar Coordinates
Rami Younes
Limits of Functions of Several Variables
Continuous Functions of Several Variables
Recommended Reading
Stewart’s Caluculus (6th edition): the titles covered in
sections 10.3, 14.2 (optional). Or equally,
Thomas’ Caluculus (13th edition): the titles covered in
sections 11.3, 14.2 (optional).
Rami Younes
Limits of Functions of Several Variables
Continuous Functions of Several Variables
Limits and Continuity
Limits Using Polar Coordinates
Definition of Limit and Continuity
Recall
Let f be a function of a real variable that is defined in an open
interval that contains a (except possibly at a):
(i)
We say lim f (x) = L, if f (x) can be made arbitrarily close
Limits
Let f be a function of two or three variables (which we denote
by x). Assume that f is defined in an open ball that contains a,
except possibly at a.
to L for all values of x chosen close enough to a.
We say lim f (x) = L, if f (x) can be made arbitrarily close to
x→a
x→a
lim f (x) may or may not exist. The limit exists if and only L for all values of x chosen close enough to a.
x→a
Continuity
if lim f (x) and lim f (x) both exist and are equal.
Let f be a function of two or three variables (which we denote
x→a+
x→a−
by x). Assume that f is defined in an open ball that contains
(iii) If f is defined in an open interval that contains a, f is said
a. f is said to be continuous at a if lim f (x) = f (a).
x→a
to be continuous at a if lim f (x) = f (a).
x→a
Important Remarks
(iv) lim |f (x)| = 0 ⇐⇒ lim f (x) = 0.
(i)
x→a
x→a
(v) Sandwich Theorem
In the plane and in space, x
If for all x near a, u(x) ≤ f (x) ≤ v (x) with
can approach a along infinitely
lim u(x) = lim v (x) = L, then lim f (x) = L.
many paths and not only two
(ii)
x→a
x→a
x→a
Remark
Limits and continuity of real functions of two or three variables can
be defined as in the above remarks (i) and (iii) except that
if f is a function of two variables then open intervals are
replaced with open disks (i.e., without their boundary circle)
if f is a function of three variables then open intervals are
replaced with solid open spheres (i.e., without their
boundary spherical shell).
we will think of intervals, disks, and spheres as balls centered
at some point with some radius.
Rami Younes
paths as in the real variable
case.
(ii)
Similar to functions of a real variable, lim f (x) exists
x→a
iff the limit exists no matter the path x −→ a and if
its value does not depend on the path x −→ a.
(iii) Functions defined using a closed formula that involves
√
basic function and algebraic operations (±, ×, ÷, n .)
are continuous at points where the formula makes
sense. Piecewise defined functions are not defined with
a closed formula.
Limits of Functions of Several Variables
Continuous Functions of Several Variables
Limits and Continuity
Limits Using Polar Coordinates
Limits and Continuity of Functions of Several Variables
Remarks
Example
Let f (x, y ) =
xy
x2 + y2
. Show that
lim
f (x, y ) does not
(i)
(x,y )→(0,0)
lim |f (x)| = 0 ⇐⇒
x→a
lim f (x) = 0, (no matter how
x→a
many variables).
exist.
(ii)
Solution
We will show that different
limits are obtained if (x, y )
approaches (0, 0) along
different paths. We choose
from the paths shown in the
figure
The sandwich theorem holds for functions of several
variables.
(iii) To prove lim f (x) does not exist, show that the the
x→a
limit deos not exist along one (well) chosen path or that
the limits along two chosen paths exist but not equal.
(iv) To prove lim f (x) exists, one often resorts to the
x→a
Let (x, y ) → (0, 0) along the x−axis, that is along y = 0,
then
sandwich theorem.
Example
f (x, y ) = f (x, 0) = 0 −
−−−−−−−
→ 0.
(x,y )→(0,0)
Let f (x, y ) =
Let (x, y ) → (0, 0) along the staright line y = x, then
f (x, y ) = f (x, x) =
x2
x2
+ x2
=
1
1
−
−−−−−−−
→ .
2 (x,y )→(0,0) 2
Since we have obtained different limits along different paths,
lim
f (x, y ) does not exist.
(x,y )→(0,0)
(
xy
x 2 +y 2
0
discontinuous at (0, 0), (why?).
if (x, y ) 6= (0, 0)
if (x, y ) 6= (0, 0)
x 2 +y 2
.
0
if (x, y ) = (0, 0)
Show that f is continuous at (0, 0).
Solution
Note that for all (x, y ) 6= (0, 0), 0 ≤
0≤
xy
2
x2 + y2
= |x|
y
y2
x 2 +y 2
≤ 1. Thus,
2
x2 + y2
≤ |x|.
As (x, y ) → (0, 0), |x| → 0. The sanwich theorem implies
that |f (x, y )| → 0 as (x, y ) → (0, 0).
Example
The function f (x, y ) =

 xy 2
is
if (x, y ) = (0, 0)
Therefore,
lim
f (x, y ) = 0 = f (0, 0) and consequently
(x,y )→(0,0)
the function is continuous at (0, 0).
Rami Younes
Limits of Functions of Several Variables
Continuous Functions of Several Variables
Limits and Continuity
Limits Using Polar Coordinates
Polar Coordinates
Remarks
(i)
Cartesian coordinates of a point P, are the directed
distances from P to 2 perpendicular axes, namely,
the x−axis and the y −axis.
(ii)
Sometimes, however, it is inconvenient to use
cartesian coordinates (for example, when studying
the gravitational pull of a point mass where it is
desired to use the radial distance from the point).
Examples
Plot the point given in polar coordinates.
Polar Coordinates
Using polar coordinates consists of locating a point with
respect to a fixed point, the pole, and a fixed initial ray
through the pole.
Remark
Contrary to cartesian coordinates, a point does not have
unique polar coordinates.
When locating a point P(r , θ)
the sign of the directed angle θ indicates the sense
of rotation of the initial ray. The initial ray is rotated
counterclockwise if θ > 0 and clockwise otherwise.
having rotated the initial ray by angle θ, the sign of
the oriented distance r determines if P lies on the
positive ray or the ray opposite to it.
Rami Younes
Limits of Functions of Several Variables
Continuous Functions of Several Variables
Limits and Continuity
Limits Using Polar Coordinates
Polar Coordinates
Equations Relating Polar and Cartesian Coordinates
Example
Find
y ln(1 + x 2 + y 2 )
lim
if it exists or show that it
From the adjacent figure,
it is easy to see that:
does not exist.
x = r cos θ
y = r sin θ
Hint: Switch to polar coordinates (r , θ) and notice that
(x, y ) −→ (0, 0) ⇐⇒ r −→ 0.
r2 = x2 + y2
tan θ =
x2 + y2
(x,y )→(0,0)
Solution
y
y ln(1 + x 2 + y 2 )
x
=
x2 + y2
Examples
In each of the following cases, graph the sets of points whose
polar coordinates satisfy the given conditions
(a)
1 ≤ r ≤ 2 and 0 ≤ θ ≤
(b)
−3 ≤ r ≤ 2 and θ = π
4
(c)
2π
3
π
2
(r sin θ) ln(1 + r 2 )
r2
= sin θ
ln(1 + r 2 )
r
Thus, for all θ and all r 6= 0, the following sandwich holds
0≤
y ln(1 + x 2 + y 2 )
x2 + y2
= | sin θ|
ln(1 + r 2 )
r
≤
However, using the Mclaurin series ln(1 + x) = x −
≤ θ ≤ 5π
6
2
1
ln(1 + r 2 ) x=r
z}|{
= lim
lim
r →0 r
r →0
r
Solution
.
2
r −
r4
2
ln(1 + r 2 )
r
x2
2
+ ...
!
+ ...
= 0.
The Sandwich Theorem thus implies that,
y ln(1 + x 2 + y 2 )
ln(1 + r 2 )
lim
= lim sin θ
= 0.
r →0
(x,y )→(0,0)
x2 + y2
r
Consequently,
lim
(x,y )→(0,0)
Rami Younes
y ln(1 + x 2 + y 2 )
x2 + y2
Limits of Functions of Several Variables
= 0.