lesson 5: continuity (section 21 handout)

Section 1.5Continuity

V63.0121.021, Calculus I

New York University

September 20, 2010

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V63.0121.021, 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.


Notes

Notes

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1

Section 1.5 : ContinuityV63.0121.021, Calculus I September 20, 2010

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


Notes

Notes

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2


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.


Notes

Notes

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3


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.


Notes

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4


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.


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 −1/4since lim

x→−1/4+f (x) = f (−1/4).


Notes

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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


Notes

Notes

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6


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


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)


Notes

Notes

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7


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.


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.


Notes

Notes

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8



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.



x

y

π

7

1

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


Notes

Notes

Notes

9


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


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.


Notes

Notes

Notes

10


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.


Notes

Notes

Notes

11


Illustrating the IVT

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


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.


Notes

Notes

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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


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.


Outline

Continuity




Notes

Notes

Notes

13



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

To be discussed in class!

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.


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).


Notes

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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


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.


Notes

Notes

Notes

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lesson 5: continuity (section 21 handout)

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