lesson 5: continuity (section 41 slides)

Section 1.5Continuity

V63.0121.041, Calculus I

New York University

September 20, 2010

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V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 2 / 47

Grader’s corner

I HW Grades will be onblackboard this week, andthe papers will be returnedin recitation

I Remember units whencomputing slopes

I Remember to staple yourpapers—you have beenwarned.


Objectives

I Understand and apply thedefinition of continuity for afunction at a point or on aninterval.

I Given a piecewise definedfunction, decide where it iscontinuous or discontinuous.

I State and understand theIntermediate ValueTheorem.

I Use the IVT to show that afunction takes a certainvalue, or that an equationhas a solution


Last time

Definition

We writelimx→a

f (x) = L

and say

“the limit of f (x), as x approaches a, equals L”

if we can make the values of f (x) arbitrarily close to L (as close to L as welike) by taking x to be sufficiently close to a (on either side of a) but notequal to a.


Limit Laws for arithmetic

Theorem (Basic Limits)

I limx→a

x = a

I limx→a

c = c

Theorem (Limit Laws)

Let f and g be functions with limits at a point a. Then

I limx→a

(f (x) + g(x)) = limx→a

f (x) + limx→a

g(x)

I limx→a

(f (x)− g(x)) = limx→a

f (x)− limx→a

g(x)

I limx→a

(f (x) · g(x)) = limx→a

f (x) · limx→a

g(x)

I limx→a

f (x)

g(x)=

limx→a f (x)

limx→a g(x)if lim

x→ag(x) 6= 0


Hatsumon

Here are some discussion questions to start.

True or False

At some point in your life you were exactly three feet tall.

True or False

At some point in your life your height (in inches) was equal to your weight(in pounds).

True or False

Right now there are a pair of points on opposite sides of the worldmeasuring the exact same temperature.


Outline

Continuity

The Intermediate Value Theorem

Back to the Questions


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

limx→a

f (x) = f (a)

This property is so useful it’s worth naming.


Definition of Continuity

Definition

I Let f be a function definednear a. We say that f iscontinuous at a if

limx→a

f (x) = f (a).

I A function f is continuousif it is continuous at everypoint in its domain. x

y

a

f (a)


Scholium

Definition

Let f be a function defined near a. We say that f is continuous at a if

limx→a

f (x) = f (a).

There are three important parts to this definition.

I The function has to have a limit at a,

I the function has to have a value at a,

I and these values have to agree.


Free Theorems

Theorem

(a) Any polynomial is continuous everywhere; that is, it is continuous onR = (−∞,∞).

(b) Any rational function is continuous wherever it is defined; that is, it iscontinuous on its domain.


Showing a function is continuous

Example

Let f (x) =√

4x + 1. Show that f is continuous at 2.

Solution

We want to show that limx→2

f (x) = f (2). We have

limx→a

f (x) = limx→2

√4x + 1 =

√limx→2

(4x + 1) =√

9 = 3 = f (2).

Each step comes from the limit laws.

Question

At which other points is f continuous?

Answer

The function f is continuous on (−1/4,∞). It is right continuous at −1/4 sincelim

x→−1/4+f (x) = f (−1/4).


At which other points?

For reference: f (x) =√

4x + 1

I If a > −1/4, then limx→a

(4x + 1) = 4a + 1 > 0, so

limx→a

f (x) = limx→a

√4x + 1 =

√limx→a

(4x + 1) =√

4a + 1 = f (a)

and f is continuous at a.

I If a = −1/4, then 4x + 1 < 0 to the left of a, which means√

4x + 1 isundefined. Still,

limx→a+

f (x) = limx→a+

√4x + 1 =

√lim

x→a+(4x + 1) =

√0 = 0 = f (a)

so f is continuous on the right at a = −1/4.



Example

Let f (x) =√


Solution



limx→a

f (x) = limx→2

√4x + 1 =

√limx→2

(4x + 1) =√

9 = 3 = f (2).


Question


Answer

The function f is continuous on (−1/4,∞).

It is right continuous at −1/4since lim

x→−1/4+f (x) = f (−1/4).



Example

Let f (x) =√


Solution



limx→a

f (x) = limx→2

√4x + 1 =

√limx→2

(4x + 1) =√

9 = 3 = f (2).


Question


Answer

The function f is continuous on (−1/4,∞). It is right continuous at −1/4since lim

x→−1/4+f (x) = f (−1/4).


The Limit Laws give Continuity Laws

Theorem

If f (x) and g(x) are continuous at a and c is a constant, then thefollowing functions are also continuous at a:

I (f + g)(x)

I (f − g)(x)

I (cf )(x)

I (fg)(x)

If

g(x) (if g(a) 6= 0)


Why a sum of continuous functions is continuous

We want to show that

limx→a

(f + g)(x) = (f + g)(a).

We just follow our nose:

limx→a

(f + g)(x) = limx→a

[f (x) + g(x)] (def of f + g)

= limx→a

f (x) + limx→a

g(x) (if these limits exist)

= f (a) + g(a) (they do; f and g are cts.)

= (f + g)(a) (def of f + g again)


Trigonometric functions are continuous

I sin and cos are continuous on R.

I tan =sin

cosand sec =

1

cosare

continuous on their domain,which isR \

{ π2

+ kπ∣∣∣ k ∈ Z

}.

I cot =cos

sinand csc =

1

sinare

continuous on their domain,which is R \ { kπ | k ∈ Z }.

sin

cos

tan sec

cot csc




I tan =sin

cosand sec =

1

cosare


{ π2

+ kπ∣∣∣ k ∈ Z

}.

I cot =cos

sinand csc =

1

sinare


sin

cos

tan

sec

cot csc




I tan =sin

cosand sec =

1

cosare


{ π2

+ kπ∣∣∣ k ∈ Z

}.

I cot =cos

sinand csc =

1

sinare


sin

cos

tan sec

cot csc




I tan =sin

cosand sec =

1

cosare


{ π2

+ kπ∣∣∣ k ∈ Z

}.

I cot =cos

sinand csc =

1

sinare


sin

cos

tan sec

cot

csc




I tan =sin

cosand sec =

1

cosare


{ π2

+ kπ∣∣∣ k ∈ Z

}.

I cot =cos

sinand csc =

1

sinare


sin

cos

tan sec

cot csc


Exponential and Logarithmic functions arecontinuous

For any base a > 1,

I the function x 7→ ax iscontinuous on R

I the function loga iscontinuous on its domain:(0,∞)

I In particular ex andln = loge are continuous ontheir domains

ax

loga x


Inverse trigonometric functions are mostlycontinuous

I sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, andright continuous at −1.

I sec−1 and csc−1 are continuous on (−∞,−1) ∪ (1,∞), leftcontinuous at −1, and right continuous at 1.

I tan−1 and cot−1 are continuous on R.

−π

−π/2

π/2

π

sin−1

cos−1 sec−1

csc−1tan−1

cot−1






−π

−π/2

π/2

π

sin−1

cos−1

sec−1

csc−1tan−1

cot−1






−π

−π/2

π/2

π

sin−1

cos−1 sec−1

csc−1tan−1

cot−1






−π

−π/2

π/2

π

sin−1

cos−1 sec−1

csc−1

tan−1

cot−1






−π

−π/2

π/2

π

sin−1

cos−1 sec−1

csc−1tan−1

cot−1


What could go wrong?

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

limx→a

f (x) = f (a)


Continuity FAIL

: The limit does not exist

Example

Let

f (x) =

{x2 if 0 ≤ x ≤ 1

2x if 1 < x ≤ 2

At which points is f continuous?

Solution

At any point a in [0, 2] besides 1, limx→a

f (x) = f (a) because f is represented by a

polynomial near a, and polynomials have the direct substitution property. However,

limx→1−

f (x) = limx→1−

x2 = 12 = 1

limx→1+

f (x) = limx→1+

2x = 2(1) = 2

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


Continuity FAIL: The limit does not exist

Example

Let

f (x) =

{x2 if 0 ≤ x ≤ 1

2x if 1 < x ≤ 2


Solution

At any point a in [0, 2] besides 1, limx→a

f (x) = f (a) because f is represented by a

polynomial near a, and polynomials have the direct substitution property. However,

limx→1−

f (x) = limx→1−

x2 = 12 = 1

limx→1+

f (x) = limx→1+

2x = 2(1) = 2

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


Graphical Illustration of Pitfall #1

x

y

−1 1 2

−1

1

2

3

4

The function cannot becontinuous at a point if thefunction has no limit at thatpoint.


Continuity FAIL

: The function has no value

Example

Let

f (x) =x2 + 2x + 1

x + 1


Solution

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


Continuity FAIL: The function has no value

Example

Let

f (x) =x2 + 2x + 1

x + 1


Solution

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



x

y

−1

1 The function cannot becontinuous at a point outside itsdomain (that is, a point where ithas no value).


Continuity FAIL

: function value 6= limit

Example

Let

f (x) =

{7 if x 6= 1

π if x = 1


Solution

f is not continuous at 1 because f (1) = π but limx→1

f (x) = 7.


Continuity FAIL: function value 6= limit

Example

Let

f (x) =

{7 if x 6= 1

π if x = 1


Solution

f is not continuous at 1 because f (1) = π but limx→1

f (x) = 7.



x

y

π

7

1

If the function has a limit and avalue at a point the two muststill agree.


Special types of discontinuites

removable discontinuity The limit limx→a

f (x) exists, but f is not defined

at a or its value at a is not equal to the limit at a.

Byre-defining f (a) = lim

x→af (x), f can be made continuous at a

jump discontinuity The limits limx→a−

f (x) and limx→a+

f (x) exist, but are

different.

The function cannot be made continuous bychanging a single value.


Graphical representations of discontinuities

x

y

π

7

1

Presto! continuous!

removable

x

y

1

1

2

continuous?

continuous?

continuous?

jump





at a or its value at a is not equal to the limit at a. Byre-defining f (a) = lim



f (x) and limx→a+


different.

The function cannot be made continuous bychanging a single value.





at a or its value at a is not equal to the limit at a. Byre-defining f (a) = lim



f (x) and limx→a+


different. The function cannot be made continuous bychanging a single value.


The greatest integer function

[[x ]] is the greatest integer ≤ x .

x [[x ]]

0 01 1

1.5 11.9 12.1 2−0.5 −1−0.9 −1−1.1 −2

x

y

−2

−2

−1

−1

1

1

2

2

3

3y = [[x ]]

This function has a jump discontinuity at each integer.


Outline

Continuity




A Big Time Theorem

Theorem (The Intermediate Value Theorem)

Suppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) 6= f (b). Then there exists anumber c in (a, b) such that f (c) = N.


Illustrating the IVT


x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3



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

and let N be anynumber between f (a) and f (b), where f (a) 6= f (b). Then there exists anumber c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3



Suppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) 6= f (b).

Then there exists anumber c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3




x

f (x)

a b

f (a)

f (b)

N

c

c1 c2 c3




x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3




x

f (x)

a b

f (a)

f (b)

N

c

c1 c2 c3


What the IVT does not say

The Intermediate Value Theorem is an “existence” theorem.

I It does not say how many such c exist.

I It also does not say how to find c.

Still, it can be used in iteration or in conjunction with other theorems toanswer these questions.


Using the IVT to find zeroes

Example

Let f (x) = x3− x − 1. Show that there is a zero for f on the interval [1, 2].

Solution

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

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


Finding a zero by bisection

x f (x)

1 − 11.25 − 0.296875

1.3125 − 0.05151371.375 0.224609

1.5 0.8752 5

(More careful analysis yields1.32472.)

x

y



x f (x)

1 − 1

1.25 − 0.2968751.3125 − 0.0515137

1.375 0.2246091.5 0.875

2 5


x

y



x f (x)

1 − 1

1.25 − 0.2968751.3125 − 0.0515137

1.375 0.224609

1.5 0.8752 5


x

y



x f (x)

1 − 11.25 − 0.296875

1.3125 − 0.05151371.375 0.224609

1.5 0.8752 5


x

y



x f (x)

1 − 11.25 − 0.296875

1.3125 − 0.0515137

1.375 0.2246091.5 0.875

2 5


x

y



x f (x)

1 − 11.25 − 0.296875

1.3125 − 0.05151371.375 0.224609

1.5 0.8752 5


x

y


Using the IVT to assert existence of numbers

Example

Suppose we are unaware of the square root function and that it’scontinuous. Prove that the square root of two exists.

Proof.

Let f (x) = x2, a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in (1, 2) suchthat

f (c) = c2 = 2.



Example


Proof.

Let f (x) = x2, a continuous function on [1, 2].

Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in (1, 2) suchthat

f (c) = c2 = 2.



Example


Proof.

Let f (x) = x2, a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in (1, 2) suchthat

f (c) = c2 = 2.


Outline

Continuity





True or False

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

True or False

At one point in your life your height in inches equaled your weight inpounds.

True or False

Right now there are two points on opposite sides of the Earth with exactlythe same temperature.


Question 1

Answer

The answer is TRUE.

I Let h(t) be height, which varies continuously over time.

I Then h(birth) < 3 ft and h(now) > 3 ft.

I So by the IVT there is a point c in (birth, now) where h(c) = 3.



True or False


True or False


True or False



Question 2

Answer

The answer is TRUE.

I Let h(t) be height in inches and w(t) be weight in pounds, bothvarying continuously over time.

I Let f (t) = h(t)− w(t).

I For most of us (call your mom), f (birth) > 0 and f (now) < 0.

I So by the IVT there is a point c in (birth, now) where f (c) = 0.

I In other words,

h(c)− w(c) = 0 ⇐⇒ h(c) = w(c).



True or False


True or False


True or False



Question 3

Answer

The answer is TRUE.

I Let T (θ) be the temperature at the point on the equator at longitudeθ.

I How can you express the statement that the temperature on oppositesides is the same?

I How can you ensure this is true?


Question 3

I Let f (θ) = T (θ)− T (θ + 180◦)

I Thenf (0) = T (0)− T (180)

whilef (180) = T (180)− T (360) = −f (0)

I So somewhere between 0 and 180 there is a point θ where f (θ) = 0!


What have we learned today?

I Definition: a function is continuous at a point if the limit of thefunction at that point agrees with the value of the function at thatpoint.

I We often make a fundamental assumption that functions we meet innature are continuous.

I The Intermediate Value Theorem is a basic property of real numbersthat we need and use a lot.


lesson 5: continuity (section 41 slides)

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