Lesson 5: Continuity

Section 1.5
Continuity

V63.0121, Calculus I

February 2–3, 2009

Announcements
Quiz 1 is this week: Tuesday (Sections 11 and 12), Thursday
(Section 4), Friday (Sections 13 and 14). 15 minutes, covers
Sections 1.1–1.2
Fill your ALEKS pie by February 27, 11:59pm
Congratulations to the Super Bowl XLIII Champion
Pittsburgh Steelers!

Outline

Warmup

Grader Notes

Continuity

The Intermediate Value Theorem

Back to the Questions

Hatsumon

Here are some discussion questions to start.
Were you ever exactly three feet tall?

Hatsumon

Was your height (in inches) ever equal to your weight (in
pounds)?

Hatsumon

Was your height (in inches) ever equal to your weight (in
pounds)?
Is there a pair of points on opposite sides of the world at the
same temperature at the same time?

But ﬁrst, a word from our friendly graders

Please turn in neat problem sets: loose-leaf paper, stapled
Label homework with Name, Section (10 or 4), Date, Problem
Set number
Do not turn in scratch work

Please label your graphs

Example: Graph F (x) = |2x + 1|.

y

(0, 1)
x
(−1/2, 0)
incomplete better

Example = explanation

Problem
If f is even and g is odd, what can you say about fg ?

Let f (x) = x 2 (even) and g (x) = x 3 (odd).
Then (fg )(x) = x 2 · x 3 = x 5 , and that’s odd.
So the product of an even function and an odd
function is an odd function.

Dangerous!

The trouble with proof by example

Problem
Which odd numbers are prime?

The numbers 3, 5, and
7 are odd, and all prime.
So all odd numbers are
prime.

Fallacious!

Use the deﬁnitions

Let f be even and g be odd. Then

(fg )(−x) = f (−x)g (−x) = f (x)(−g (x))
= −f (x)g (x) = −(fg )(x)

So fg is odd.

Better

Recall: Direct Substitution Property

Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of
f , then
lim f (x) = f (a)
x→a

Definition of Continuity

Definition
Let f be a function defined near a. We say that f is continuous at
a if
lim f (x) = f (a).
x→a

Free Theorems

Theorem
(a) Any polynomial is continuous everywhere; that is, it is
continuous on R = (−∞, ∞).
(b) Any rational function is continuous wherever it is deﬁned; that
is, it is continuous on its domain.

Showing a function is continuous

Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.


Example
√

Solution
We have
√
lim f (x) = lim 4x + 1
x→a x→2

= lim (4x + 1)
x→2
√
= 9 = 3.

Each step comes from the limit laws.


Example
√

Solution
We have
√
x→a x→2

= lim (4x + 1)
x→2
√
= 9 = 3.

In fact, f is continuous on (−1/4, ∞).


Example
√

Solution
We have
√
x→a x→2

= lim (4x + 1)
x→2
√
= 9 = 3.

In fact, f is continuous on (−1/4, ∞). The function f is right
continuous at the point −1/4.

The Limit Laws give Continuity Laws

Theorem
If f and g are continuous at a and c is a constant, then the
following functions are also continuous at a:
f +g
f −g
cf
fg
f
(if g (a) = 0)
g

Transcendental functions are continuous, too

Theorem
The following functions are continuous wherever they are deﬁned:
1. sin, cos, tan, cot sec, csc
2. x → ax , loga , ln
3. sin−1 , tan−1 , sec−1

What could go wrong?

In what ways could a function f fail to be continuous at a point a?
Look again at the deﬁnition:

lim f (x) = f (a)
x→a

Pitfall #1
Example
Let
x2 if 0 ≤ x ≤ 1
f (x) =
if 1 < x ≤ 2
2x
At which points is f continuous?

Pitfall #1: The limit does not exist
Example
Let
x2 if 0 ≤ x ≤ 1
f (x) =
if 1 < x ≤ 2
2x

Solution
At any point a in [0, 2] besides 1, lim f (x) = f (a) because f is
x→a
represented by a polynomial near a, and polynomials have the
direct substitution property. However,

lim f (x) = lim x 2 = 12 = 1
x→1− x→1−
lim f (x) = lim+ 2x = 2(1) = 2
x→1+ x→1

So f has no limit at 1. Therefore f is not continuous at 1.

Graphical Illustration of Pitfall #1

y
4

3

2

1

x
−1 1 2
−1

Pitfall #2

Example
Let
x 2 + 2x + 1
f (x) =
x +1

Pitfall #2: The function has no value

Example
Let
x 2 + 2x + 1
f (x) =
x +1

Solution
Because f is rational, it is continuous on its whole domain. Note
that −1 is not in the domain of f , so f is not continuous there.


y

1

x
−1

f cannot be continuous where it has no value.

Pitfall #3

Example
Let
7 if x = 1
f (x) =
π if x = 1

Pitfall #3: function value = limit

Example
Let
7 if x = 1
f (x) =
π if x = 1

Solution
f is not continuous at 1 because f (1) = π but lim f (x) = 7.
x→1


y

7

π

x
1

Special types of discontinuites

removable discontinuity The limit lim f (x) exists, but f is not
x→a
deﬁned at a or its value at a is not equal to the limit
at a.
jump discontinuity The limits lim f (x) and lim+ f (x) exist, but
x→a− x→a
are diﬀerent. f (a) is one of these limits.

Graphical representations of discontinuities

y
4
y
3
7
2
π
1
x
x
1
−1 1 2
removable −1
jump

The greatest integer function

y

3

2

1

x
−2 −1 1 2 3
−1

−2

The greatest integer function

y

3

2

1

x
−2 −1 1 2 3
−1

−2

The greatest integer function f (x) = [[x]] has jump discontinuities.

A Big Time Theorem

Theorem (The Intermediate Value Theorem)
Suppose that f is continuous on the closed interval [a, b] and let N
be any number between f (a) and f (b), where f (a) = f (b). Then
there exists a number c in (a, b) such that f (c) = N.

Illustrating the IVT

f (x)

x

Suppose that f is continuous on the closed interval [a, b]

f (x)

x

Suppose that f is continuous on the closed interval [a, b]

f (x)

f (b)

f (a)

x
a b

be any number between f (a) and f (b), where f (a) = f (b).

f (x)

f (b)

N

f (a)

x
a b

f (x)

f (b)

N

f (a)

x
a c b

f (x)

f (b)

N

f (a)

x
a b

f (x)

f (b)

N

f (a)

x
a c1 c2 c3 b

Using the IVT

Example
Prove that the square root of two exists.

Using the IVT

Example

Proof.
Let f (x) = x 2 , a continuous function on [1, 2].

Using the IVT

Example

Proof.
Let f (x) = x 2 , a continuous function on [1, 2]. Note f (1) = 1 and
f (2) = 4. Since 2 is between 1 and 4, there exists a point c in
(1, 2) such that
f (c) = c 2 = 2.

Using the IVT

Example

Proof.
Let f (x) = x 2 , a continuous function on [1, 2]. Note f (1) = 1 and
f (2) = 4. Since 2 is between 1 and 4, there exists a point c in
(1, 2) such that
f (c) = c 2 = 2.

In fact, we can “narrow in” on the square root of 2 by the method
of bisections.

√
Finding 2 by bisections

x f (x) = x 2

24

11

√

x f (x) = x 2

24

1.5 2.25

11

√

x f (x) = x 2

24

1.5 2.25

1.25 1.5625

11

√

x f (x) = x 2

24

1.5 2.25
1.375 1.890625
1.25 1.5625

11

√

x f (x) = x 2

24

1.5 2.25
1.4375 2.06640625
1.375 1.890625
1.25 1.5625

11

Using the IVT

Example
Let f (x) = x 3 − x − 1. Show that there is a zero for f .

Solution
f (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.

Using the IVT

Example
Let f (x) = x 3 − x − 1. Show that there is a zero for f .

Solution
f (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.
(More careful analysis yields 1.32472.)


True or False
At one point in your life you were exactly three feet tall.

Question 1: True!

Let h(t) be height, which varies continuously over time. Then
h(birth) < 3 ft and h(now) > 3 ft. So there is a point c in
(birth, now) where h(c) = 3.


True or False

True or False
At one point in your life your height in inches equaled your weight
in pounds.

Question 2: True!

Let h(t) be height in inches and w (t) be weight in pounds, both
varying continuously over time. Let f (t) = h(t) − w (t). For most
of us (call your mom), f (birth) > 0 and f (now) < 0. So there is a
point c in (birth, now) where f (c) = 0. In other words,

h(c) − w (c) = 0 ⇐⇒ h(c) = w (c).


True or False

True or False
At one point in your life your height in inches equaled your weight
in pounds.

True or False
Right now there are two points on opposite sides of the Earth with
exactly the same temperature.

Question 3

Let T (θ) be the temperature at the point on the equator at
longitude θ.
How can you express the statement that the temperature on
opposite sides is the same?
How can you ensure this is true?

Lesson 5: Continuity

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