Lesson 32 – Continuity of Functions Calculus - Santowski 10/13/20151Calculus - Santowski.
Lesson 4: Continuity
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Section 1.5Continuity
V63.0121.002.2010Su, Calculus I
New York University
May 20, 2010
Announcements
I Office Hours: MR 5:00–5:45, TW 7:50–8:30, CIWW 102 (here)I Quiz 1 Thursday on 1.1–1.4
. . . . . .
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. . . . . .
Announcements
I Office Hours: MR5:00–5:45, TW 7:50–8:30,CIWW 102 (here)
I Quiz 1 Thursday on1.1–1.4
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 2 / 46
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. . . . . .
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 ordiscontinuous.
I State and understand theIntermediate ValueTheorem.
I Use the IVT to show that afunction takes a certainvalue, or that an equationhas a solution
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 3 / 46
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. . . . . .
Last time
DefinitionWe write
limx→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 aswe like) by taking x to be sufficiently close to a (on either side of a) butnot equal to a.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 4 / 46
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. . . . . .
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. ThenI lim
x→a(f(x) + g(x)) = lim
x→af(x) + lim
x→ag(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 limx→a
g(x) ̸= 0
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 5 / 46
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. . . . . .
Hatsumon
Here are some discussion questions to start.
True or FalseAt some point in your life you were exactly three feet tall.
True or FalseAt some point in your life your height (in inches) was equal to yourweight (in pounds).
True or FalseRight now there are a pair of points on opposite sides of the worldmeasuring the exact same temperature.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 6 / 46
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. . . . . .
Hatsumon
Here are some discussion questions to start.
True or FalseAt some point in your life you were exactly three feet tall.
True or FalseAt some point in your life your height (in inches) was equal to yourweight (in pounds).
True or FalseRight now there are a pair of points on opposite sides of the worldmeasuring the exact same temperature.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 6 / 46
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. . . . . .
Hatsumon
Here are some discussion questions to start.
True or FalseAt some point in your life you were exactly three feet tall.
True or FalseAt some point in your life your height (in inches) was equal to yourweight (in pounds).
True or FalseRight now there are a pair of points on opposite sides of the worldmeasuring the exact same temperature.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 6 / 46
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. . . . . .
Outline
Continuity
The Intermediate Value Theorem
Back to the Questions
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 7 / 46
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. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 8 / 46
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. . . . . .
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 9 / 46
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. . . . . .
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 9 / 46
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. . . . . .
Scholium
DefinitionLet 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 10 / 46
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. . . . . .
Free Theorems
Theorem
(a) Any polynomial is continuous everywhere; that is, it is continuouson R = (−∞,∞).
(b) Any rational function is continuous wherever it is defined; that is, itis continuous on its domain.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 11 / 46
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. . . . . .
Showing a function is continuous.
.
Example
Let f(x) =√4x+ 1. Show that f is continuous at 2.
SolutionWe want to show that lim
x→2f(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.
QuestionAt which other points is f continuous?
AnswerThe function f is continuous on (−1/4,∞). It is right continuous at −1/4since lim
x→−1/4+f(x) = f(−1/4).
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 12 / 46
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. . . . . .
Showing a function is continuous.
.
Example
Let f(x) =√4x+ 1. Show that f is continuous at 2.
SolutionWe want to show that lim
x→2f(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.
QuestionAt which other points is f continuous?
AnswerThe function f is continuous on (−1/4,∞). It is right continuous at −1/4since lim
x→−1/4+f(x) = f(−1/4).
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 12 / 46
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. . . . . .
Showing a function is continuous.
.
Example
Let f(x) =√4x+ 1. Show that f is continuous at 2.
SolutionWe want to show that lim
x→2f(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.
QuestionAt which other points is f continuous?
AnswerThe function f is continuous on (−1/4,∞). It is right continuous at −1/4since lim
x→−1/4+f(x) = f(−1/4).
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 12 / 46
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. . . . . .
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
is undefined. 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 13 / 46
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. . . . . .
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
is undefined. 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 13 / 46
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. . . . . .
Showing a function is continuous
Example
Let f(x) =√4x+ 1. Show that f is continuous at 2.
SolutionWe want to show that lim
x→2f(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.
QuestionAt which other points is f continuous?
AnswerThe function f is continuous on (−1/4,∞).
It is right continuous at−1/4 since lim
x→−1/4+f(x) = f(−1/4).
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 14 / 46
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. . . . . .
Showing a function is continuous
Example
Let f(x) =√4x+ 1. Show that f is continuous at 2.
SolutionWe want to show that lim
x→2f(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.
QuestionAt which other points is f continuous?
AnswerThe function f is continuous on (−1/4,∞). It is right continuous at−1/4 since lim
x→−1/4+f(x) = f(−1/4).V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 14 / 46
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. . . . . .
The Limit Laws give Continuity Laws
TheoremIf 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)
Ifg(x) (if g(a) ̸= 0)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 15 / 46
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. . . . . .
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 16 / 46
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. . . . . .
Trigonometric functions are continuous
I sin and cos are continuous onR.
I tan =sincos
and sec =1cos
arecontinuous on their domain,which isR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuous on their domain,which is R \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot .csc
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
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. . . . . .
Trigonometric functions are continuous
I sin and cos are continuous onR.
I tan =sincos
and sec =1cos
arecontinuous on their domain,which isR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuous on their domain,which is R \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot .csc
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
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. . . . . .
Trigonometric functions are continuous
I sin and cos are continuous onR.
I tan =sincos
and sec =1cos
arecontinuous on their domain,which isR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuous on their domain,which is R \ { kπ | k ∈ Z }.
..sin
.cos
.tan
.sec
.cot .csc
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
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. . . . . .
Trigonometric functions are continuous
I sin and cos are continuous onR.
I tan =sincos
and sec =1cos
arecontinuous on their domain,which isR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuous on their domain,which is R \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot .csc
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
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. . . . . .
Trigonometric functions are continuous
I sin and cos are continuous onR.
I tan =sincos
and sec =1cos
arecontinuous on their domain,which isR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuous on their domain,which is R \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot
.csc
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
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. . . . . .
Trigonometric functions are continuous
I sin and cos are continuous onR.
I tan =sincos
and sec =1cos
arecontinuous on their domain,which isR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuous on their domain,which is R \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot .csc
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
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. . . . . .
Exponential and Logarithmic functions are continuous
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 continuouson their domains
.
.ax
.loga x
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 18 / 46
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. . . . . .
Exponential and Logarithmic functions are continuous
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 continuouson their domains
.
.ax
.loga x
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 18 / 46
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. . . . . .
Exponential and Logarithmic functions are continuous
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 continuouson their domains
.
.ax
.loga x
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 18 / 46
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. . . . . .
Inverse trigonometric functions are mostly continuous
I sin−1 and cos−1 are continuous on (−1,1), left continuous at 1,and right 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−1
.
.
.tan−1
.cot−1
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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. . . . . .
Inverse trigonometric functions are mostly continuous
I sin−1 and cos−1 are continuous on (−1,1), left continuous at 1,and right 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−1
.
.
.tan−1
.cot−1
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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. . . . . .
Inverse trigonometric functions are mostly continuous
I sin−1 and cos−1 are continuous on (−1,1), left continuous at 1,and right 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−1
.
.
.tan−1
.cot−1
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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. . . . . .
Inverse trigonometric functions are mostly continuous
I sin−1 and cos−1 are continuous on (−1,1), left continuous at 1,and right 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−1
.
.
.tan−1
.cot−1
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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. . . . . .
Inverse trigonometric functions are mostly continuous
I sin−1 and cos−1 are continuous on (−1,1), left continuous at 1,and right 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−1
.
.
.tan−1
.cot−1
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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. . . . . .
Inverse trigonometric functions are mostly continuous
I sin−1 and cos−1 are continuous on (−1,1), left continuous at 1,and right 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−1
.
.
.tan−1
.cot−1
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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. . . . . .
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 20 / 46
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. . . . . .
Continuity FAIL
: The limit does not exist
.
.
Example
Let
f(x) =
{x2 if 0 ≤ x ≤ 12x if 1 < x ≤ 2
At which points is f continuous?
SolutionAt any point a in [0,2] besides 1, lim
x→af(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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 21 / 46
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. . . . . .
Continuity FAIL: The limit does not exist.
.
Example
Let
f(x) =
{x2 if 0 ≤ x ≤ 12x if 1 < x ≤ 2
At which points is f continuous?
SolutionAt any point a in [0,2] besides 1, lim
x→af(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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 21 / 46
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. . . . . .
Graphical Illustration of Pitfall #1
. .x
.y
..−1
..1
..2
..−1
..1
..2
..3
..4
.
.
.
.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 22 / 46
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. . . . . .
Continuity FAIL
: The function has no value
Example
Let
f(x) =x2 + 2x+ 1
x+ 1At which points is f continuous?
SolutionBecause 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 23 / 46
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. . . . . .
Continuity FAIL: The function has no value
Example
Let
f(x) =x2 + 2x+ 1
x+ 1At which points is f continuous?
SolutionBecause 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 23 / 46
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. . . . . .
Graphical Illustration of Pitfall #2
. .x
.y
...−1
. .1
f cannot be continuous where it has no value.
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. . . . . .
Continuity FAIL
: function value ̸= limit
Example
Let
f(x) =
{7 if x ̸= 1π if x = 1
At which points is f continuous?
Solutionf is not continuous at 1 because f(1) = π but lim
x→1f(x) = 7.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 25 / 46
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. . . . . .
Continuity FAIL: function value ̸= limit
Example
Let
f(x) =
{7 if x ̸= 1π if x = 1
At which points is f continuous?
Solutionf is not continuous at 1 because f(1) = π but lim
x→1f(x) = 7.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 25 / 46
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. . . . . .
Graphical Illustration of Pitfall #3
. .x
.y
..π
..7
..1
.
.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 26 / 46
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. . . . . .
Special types of discontinuites
removable discontinuity The limit limx→a
f(x) exists, but f is not definedat 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 aredifferent.
The function cannot be made continuous bychanging a single value.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 27 / 46
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. . . . . .
Graphical representations of discontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?.
. .continuous?
. .continuous?
jump
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
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. . . . . .
Graphical representations of discontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?.
. .continuous?
. .continuous?
jump
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
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. . . . . .
Graphical representations of discontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?
.
. .continuous?
. .continuous?
jump
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
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. . . . . .
Graphical representations of discontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?
.
. .continuous?
. .continuous?
jump
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
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. . . . . .
Graphical representations of discontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?
.
. .continuous?
. .continuous?
jump
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
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. . . . . .
Special types of discontinuites
removable discontinuity The limit limx→a
f(x) exists, but f is not definedat 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 aredifferent.
The function cannot be made continuous bychanging a single value.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 29 / 46
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. . . . . .
Special types of discontinuites
removable discontinuity The limit limx→a
f(x) exists, but f is not definedat 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 aredifferent. The function cannot be made continuous bychanging a single value.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 29 / 46
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. . . . . .
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
..3
. .
. .
. .
. .
. ..y = [[x]]
This function has a jump discontinuity at each integer.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 30 / 46
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. . . . . .
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
..3
. .
. .
. .
. .
. ..y = [[x]]
This function has a jump discontinuity at each integer.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 30 / 46
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. . . . . .
Outline
Continuity
The Intermediate Value Theorem
Back to the Questions
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 31 / 46
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. . . . . .
A Big Time Theorem
Theorem (The Intermediate Value Theorem)
Suppose that f is continuous on the closed interval [a,b] and let N beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 32 / 46
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. . . . . .
Illustrating the IVT
Suppose that f is continuous on the closed interval [a,b] and let N beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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. . . . . .
Illustrating the IVT
Suppose that f is continuous on the closed interval [a,b]
and let N beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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. . . . . .
Illustrating the IVT
Suppose that f is continuous on the closed interval [a,b]
and let N beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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. . . . . .
Illustrating the IVT
Suppose that f is continuous on the closed interval [a,b] and let N beany number between f(a) and f(b), where f(a) ̸= f(b).
Then thereexists a number c in (a,b) such that f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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. . . . . .
Illustrating the IVT
Suppose that f is continuous on the closed interval [a,b] and let N beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
.
.
.c1
.
.c2
.
.c3
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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. . . . . .
Illustrating the IVT
Suppose that f is continuous on the closed interval [a,b] and let N beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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. . . . . .
Illustrating the IVT
Suppose that f is continuous on the closed interval [a,b] and let N beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
.
.
.c1
.
.c2
.
.c3V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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. . . . . .
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.
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. . . . . .
Using the IVT
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)such that
f(c) = c2 = 2.
In fact, we can “narrow in” on the square root of 2 by the method ofbisections.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 35 / 46
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. . . . . .
Using the IVT
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)such that
f(c) = c2 = 2.
In fact, we can “narrow in” on the square root of 2 by the method ofbisections.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 35 / 46
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. . . . . .
Using the IVT
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)such that
f(c) = c2 = 2.
In fact, we can “narrow in” on the square root of 2 by the method ofbisections.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 35 / 46
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. . . . . .
Using the IVT
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)such that
f(c) = c2 = 2.
In fact, we can “narrow in” on the square root of 2 by the method ofbisections.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 35 / 46
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. . . . . .
Finding√2 by bisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
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. . . . . .
Finding√2 by bisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
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. . . . . .
Finding√2 by bisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
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. . . . . .
Finding√2 by bisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
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. . . . . .
Finding√2 by bisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
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. . . . . .
Finding√2 by bisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
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. . . . . .
Using the IVT
Example
Let f(x) = x3 − x− 1. Show that there is a zero for f.
Solutionf(1) = −1 and f(2) = 5. So there is a zero between 1 and 2.
(Morecareful analysis yields 1.32472.)
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. . . . . .
Using the IVT
Example
Let f(x) = x3 − x− 1. Show that there is a zero for f.
Solutionf(1) = −1 and f(2) = 5. So there is a zero between 1 and 2. (Morecareful analysis yields 1.32472.)
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. . . . . .
Outline
Continuity
The Intermediate Value Theorem
Back to the Questions
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. . . . . .
Back to the Questions
True or FalseAt one point in your life you were exactly three feet tall.
True or FalseAt one point in your life your height in inches equaled your weight inpounds.
True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature.
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. . . . . .
Question 1: 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.
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. . . . . .
Back to the Questions
True or FalseAt one point in your life you were exactly three feet tall.
True or FalseAt one point in your life your height in inches equaled your weight inpounds.
True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature.
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. . . . . .
Question 2: 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).
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. . . . . .
Back to the Questions
True or FalseAt one point in your life you were exactly three feet tall.
True or FalseAt one point in your life your height in inches equaled your weight inpounds.
True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature.
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. . . . . .
Question 3
I Let T(θ) be the temperature at the point on the equator atlongitude θ.
I How can you express the statement that the temperature onopposite sides is the same?
I How can you ensure this is true?
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. . . . . .
Question 3: True!
I Let f(θ) = T(θ)− T(θ + 180◦)I Then
f(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 wheref(θ) = 0!
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. . . . . .
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 meetin nature are continuous.
I The Intermediate Value Theorem is a basic property of realnumbers that we need and use a lot.
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