Recall the following properties of limits.
Theorem. Suppose that lim f (x) = L and lim g(x) = M . Then
x→a
x→a
• lim [f (x) ± g(x)] = L + M ,
x→a
• lim [f (x)g(x)] = LM ,
x→a
• if M �= 0, lim
x→a
f (x)
L
= .
g(x) M
Furthermore, if f (x) ≤ g(x) for all x, then L ≤ M .
Another useful result on limits is the squeeze, or sandwich, theorem:
Theorem. Suppose f, g, h are functions such that for all x (except
possibly at a)
g(x) ≤ f (x) ≤ h(x).
Also assume that
lim g(x) = lim h(x) = L.
x→a
x→a
Then
lim f (x) = L.
x→a
See slide 1 for example that utilizes the squeeze theorem.
Continuity
Definition A function f (x) is continuous at a if
lim f (x) = f (a).
x→a
There are 3 implications of this def’n:
(1) the limit lim f (x) exists,
x→a
Math 30G - Prof. Kindred - Lecture 3
Page 1
(2) f (a) is defined,
(3) and the two above quantities are equal.
Example Show that f (x) = 5x − 3 is continuous at x = 1.
Last class, we showed that lim (5x − 3) = 2 via an � − δ proof. Furtherx→1
more, we know that f (1) = 2, so it follows that
lim (5x − 3) = 2 = f (1).
x→1
Thus, f (x) is continuous at x = 1.
Question How can a function fail to be continuous at a point a?
• the limit at the point a could fail to exist
(see examples of how limits fail to exist from last class)
• the function may be undefined at that point (f (a) undefined)
• lim f (x) �= f (a)
x→a
sin θ
θ
Example Is f (θ) =
continuous at θ = 0?
1
No – since f (0) is undefined.
0.8
f(!)
0.6
0.4
0.2
0
−0.2
−6π −5π −4π −3π −2π −π
0
!
π
sin θ
= 1.
θ→0 θ
Claim: lim
Math 30G - Prof. Kindred - Lecture 3
2π
3π
4π
5π
6π
But we can remove this
discontinuity because
sin θ
lim
exists!
θ→0 θ
We can’t use l’Hopital’s rule, as we
have not proven it yet.
Page 2
Proof. Assume 0 < θ <
π
. (This is OK since we are letting θ go to 0.)
2
Consider the following regions and their areas:
(1, tan θ)
(cos θ, sin θ)
(cos θ, sin θ)
θ
θ
θ
1
1
1
We have
1
sin θ cos θ
2
θ
(π · 12)
2π
≤
Then, by multiplying by
≤
1
(1) tan θ.
2
2
, we get
sin θ
θ
1
≤
sin θ
cos θ
1
sin θ
=⇒
≥
≥ cos θ.
cos θ
θ
We let θ → 0+, so then by the squeeze theorem,
sin θ
1 ≥ lim
≥ 1.
θ→0+ θ
cos θ
≤
Thus, limθ→0+ sinθ θ = 1. Notice that sinθ θ is an even function,1 so we
conclude that
sin θ
lim
= 1.
θ→0 θ
�
Now, we define
F (θ) =
�
sin θ
θ
1
if θ �= 0
if θ = 0.
1
An even function is a function f such that f (−x) = f (x) for all x. The graph of an even function is symmetric
with respect to the y-axis.
Math 30G - Prof. Kindred - Lecture 3
Page 3
F (θ) is continuous at θ = 0. Furthermore, F (θ) is continuous at all
θ �= 0 since it is the ratio of two continuous functions.
The function F (θ) is sometimes called a “continuous extension” of f (θ).
Definition We say f (x) is continuous on a set S if f (x) is continuous
at every x ∈ S.
Important properties of continuous functions
• sums and products of continuous functions are continuous
(follows from limit properties)
• f (x) = x is continuous and f (x) = c for some constant c is continuous (verify yourself as an exercise)
• using the two above properties, we know that any polynomial is a
continuous function
Theorem (extreme value theorem). If f is a continuous function on
interval [a, b], then
f has an absolute maximum at some value xmax ∈ [a, b]
and f has an absolute minimum at some value xmin ∈ [a, b].
y
y
y
f(x)
f(x)
f(x)
a
b
xmax
x
xmin
Math 30G - Prof. Kindred - Lecture 3
a
b
x
a
x
b
f continuous but only on (a, b)
f not continuous on [a, b]
no max or min on this interval
no max or min on this interval
Page 4
Theorem (intermediate value theorem). If f is a continuous function
on [a, b] and y0 is any value between f (a) and f (b), then there exists
a number c ∈ [a, b] such that f (c) = y0.
y
f(x)
y0
a
b
x
The theorem seems fairly
obvious, but it is not so
easy to prove
(Math 131 - Analysis).
c
Note that the theorem guarantees that such a number c exists without
stating how to find it.
An important corollary follows...
Corollary (existence of roots). If f is a continuous function on [a, b]
and one of f (a) and f (b) is positive and other is negative, then there
exists a value c ∈ [a, b] such that f (c) = 0.
Proof. Since f is continuous and 0 is between f (a) and f (b), it follows
from the intermediate value theorem that there exists at least one value
c ∈ (a, b) such that f (c) = 0.
�
Math 30G - Prof. Kindred - Lecture 3
Page 5
Preservation of continuity – the theorem below formalizes our previous
statement that sums and products of continuous functions are continuous.
Theorem. Assume that f (x) and g(x) are continuous at point a.
Then the following functions are also continuous at point a:
• f (x) ± g(x)
• f (x)g(x),
•
f (x)
if g(a) �= 0.
g(x)
Math 30G - Prof. Kindred - Lecture 3
Page 6
Continuity
Math 30G, Calculus
Professor Kindred
September 10, 2012
Example using squeeze/sandwich theorem
� �
1
.
We want to determine lim x sin
x→0
x
2
We cannot use the property that
lim f (x)g(x) = lim f (x) · lim g(x)
x→a
� �x→a
1
does not exist.
since lim sin
x→0
x
x→a
However, we do know that
� �
1
≤ 1 for all x �= 0
−1 ≤ sin
x
� �
1
2
2
≤ x 2 for all x �= 0
⇒ −x ≤ x sin
x
and since lim −x 2 = lim x 2 = 0,
� �
1
= 0.
by the squeeze theorem, it must be that lim x sin
x→0
x
x→0
x→0
2