Limits and Continuity: Motivation, Highlights ...cfcid/Classes/1441/Supplements/limtsandcontinu… · Limits and Continuity: Motivation, Highlights, Illustrative Problems Charles

Limits andContinuity:Motivation,Highlights,IllustrativeProblems

CharlesDelman

Limits & Area

Limits, Slopes& ExtremeValues

Evaluationand Definitionof Limits

LimitTheorems

Limits ofAlgebraicCombinations

Types ofFunctions

The SqueezeTheorem

Continuity

Limits and Continuity:Motivation, Highlights, Illustrative Problems

Charles Delman

August 23, 2013


CharlesDelman

Limits & Area



LimitTheorems


Types ofFunctions

The SqueezeTheorem

Continuity

1 Limits & Area

2 Limits, Slopes & Extreme Values

3 Evaluation and Definition of Limits

4 Limit TheoremsLimits of Algebraic CombinationsTypes of FunctionsThe Squeeze Theorem

5 Continuity


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Estimating the Area of a Circle

r

The area of a circle is clearly proportional to the square ofits radius.

That is, A = kr 2.

Clearly, k < 4. Why?

And k > 2. Why?


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In Fact, Dissection of the Regular DodecagonShows that k is Slightly Greater than 3.


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Computing the Area of a Circle Exactly

In fact, k = π (as you probably remember).

Remember that π is defined in terms of linearmeasurements; it is the ratio of circumference to diameter.

Thus, we have another deep relationship between lengthand area!

C

2r= π =

A

r 2

Why does π, the ratio of circumference to diameter, alsoturn out to be the ratio of the area of the circle to thearea of a square on the radius?

Is it just a miracle, or can we understand the reason?


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The Circled is Filled by Inscribed Polygonswith Increasing Numbers of Sides

h b

As the number of sides, n, increases, the area of theinscribed n-gon approaches the area of the circle.


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The Area of a Triangle

h b

Each triangle has area1

2bh.


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The Area of the Whole Regular N-gon

h b

There are n triangles.

So the area of the inscribed polygon isn

2bh.


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Relating Area to Perimeter; Computing the Limit

h

nb

nb is the perimeter of the polygon.

As n→∞, nb → C , the circumference of the circle, andh→ r , the radius of the circle. Remember that C = 2πr .


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Why A = πr 2: Conclusion

Thus, as n→∞, the area of the inscribed polygon,(nb)h

2, approaches

2πr · r2

= πr 2.


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Extreme Values of f (x) = (x − 1)(x − 2)(x − 3)

5

4

3

2

1

-1

-2

-3

-4

-5

-6 -4 -2 2 4 6

y=

y

x

f x( ) = x-1( )⋅ x-2( )⋅ x-3( )

The function f described by the equationf (x) = (x − 1)(x − 2)(x − 3) has two extreme values.

It has a local maximum between x = 1 and x = 2.

It has a local minimum between x = 2 and x = 3.


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The Slope at an Extreme Value is Zero

2.5

2

1.5

1

0.5

-0.5

-1

-1.5

-2

-1 1 2 3 4 5

y=

y

x

f x( ) = x-1( )⋅ x-2( )⋅ x-3( )

The curve has a slope at every point.

At the extreme values, the slope of the curve is zero.

But what does the slope of a curve mean, exactly?

How do we compute it?


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The Slope of a Line

Δx = x2 - x1

Δy = y2 - y1

(x2,y2)

(x1,y1)

Recall that the slope of a line is defined to be ratio ∆y∆x ,

where ∆y is the change in y and ∆x is the change in x .

Think of this ratio as the rate at which y changes withrespect to x .

To compute these changes, we must compare two pointson the line.

How to compute the slope of a curve at a single point?


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The Slope of a Curve is a Limit

0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6

ΔyΔx =

f(x+Δx)-f(x)Δx

(x+Δx, f(x+Δx)(x, f(x))

x+ΔxΔx

x

To find the slope of the curve, we consider another inputvalue at distance ∆x from x .

The slope of the curve at(x , f (x)

)is

lim∆x→0

f (x + ∆x)− f (x)

∆x.

Think of the slope of the curve at a point as the rate atwhich y is changing with respect to x at that point.


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Interpreting the Slope of a Curve

An interpretation assigns meaning to the variables.

For example, suppose s is defined to be the displacementof a particle (from some starting point), in meters, t isdefined to be time, in seconds, and the function f givesthe displacement at each time (s = f (t)).

The slope between(t, f (t)

)and

(t + ∆t, f (t + ∆t)

)is

the average velocity of the particle, in meters per second( m

s ), between times t and t + ∆t.

The slope of the curve s = f (t) at the point(t, f (t)

)is

the instantaneous velocity of the particle, in meters persecond ( m

s ), at time t (that is, at that instant).

Note that the units make sense: ∆s is in meters; ∆t is inseconds; thus, the rate ∆s

∆t is in meters per second; thelimit has the same units.


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Exercises: Graphing, Slopes, & Extrema

1 Graph the curve given by the equation

y = (x − 1)(x − 2).

2 Compute a formula for the slope of this curve at any point.

3 For what value of x is the value of y on the curve at itsminimum? Is this what you expected? Why?


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The Concept of Limit is very General

Note that the limiting process for finding the slope of acurve is continuous rather than sequential, as was the casefor the polygons inscribed in a circle. (The value of ∆xvaries continuously rather than approaching its limitingvalue step-by-step.)

There are many different types of limits.

Some of them are tricky to compute correctly.

So let’s develop intuition by exploring some examples.


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Exercises: Evaluate the Limits

1 What is limx→0sin xx ?

2 What is limx→0 x2?

3 What is limx→0x3

x ?

4 What is limx→0 10x2?

5 What is limx→0 100x2?. . .

6 Is limx→0 sin( 1x ) = 0?

7 Is limx→0(.1) sin( 1x ) = 0?

8 Is limx→0(.01) sin( 1x ) = 0?

. . .

9 Is limx→0 x sin( 1x ) = 0?


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1 What is limx→0sin xx ? 1

2 What is limx→0 x2?


x ?


5 What is limx→0 100x2?. . .

6 Is limx→0 sin( 1x ) = 0?

7 Is limx→0(.1) sin( 1x ) = 0?

8 Is limx→0(.01) sin( 1x ) = 0?

. . .

9 Is limx→0 x sin( 1x ) = 0?


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2 What is limx→0 x2? 0


x ?


5 What is limx→0 100x2?. . .

6 Is limx→0 sin( 1x ) = 0?

7 Is limx→0(.1) sin( 1x ) = 0?

8 Is limx→0(.01) sin( 1x ) = 0?

. . .

9 Is limx→0 x sin( 1x ) = 0?


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x ? 0


5 What is limx→0 100x2?. . .

6 Is limx→0 sin( 1x ) = 0?

7 Is limx→0(.1) sin( 1x ) = 0?

8 Is limx→0(.01) sin( 1x ) = 0?

. . .

9 Is limx→0 x sin( 1x ) = 0?


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x ? 0

4 What is limx→0 10x2? 0

5 What is limx→0 100x2?. . .

6 Is limx→0 sin( 1x ) = 0?

7 Is limx→0(.1) sin( 1x ) = 0?

8 Is limx→0(.01) sin( 1x ) = 0?

. . .

9 Is limx→0 x sin( 1x ) = 0?


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x ? 0


5 What is limx→0 100x2? 0. . .

6 Is limx→0 sin( 1x ) = 0?

7 Is limx→0(.1) sin( 1x ) = 0?

8 Is limx→0(.01) sin( 1x ) = 0?

. . .

9 Is limx→0 x sin( 1x ) = 0?


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x ? 0


5 What is limx→0 100x2? 0. . .

6 Is limx→0 sin( 1x ) = 0? No.

sin( 1x ) does not approach any limit as x → 0. The

specified limit does not exist.

7 Is limx→0(.1) sin( 1x ) = 0?

8 Is limx→0(.01) sin( 1x ) = 0?

. . .

9 Is limx→0 x sin( 1x ) = 0?


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x ? 0


5 What is limx→0 100x2? 0. . .


sin( 1x ) does not approach any limit as x → 0.

The specified limit does not exist.

7 Is limx→0(.1) sin( 1x ) = 0? No. It does not exist.

8 Is limx→0(.01) sin( 1x ) = 0?

. . .

9 Is limx→0 x sin( 1x ) = 0?


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x ? 0


5 What is limx→0 100x2? 0. . .






. . .

9 Is limx→0 x sin( 1x ) = 0?


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x ? 0


5 What is limx→0 100x2? 0. . .






. . .

9 Is limx→0 x sin( 1x ) = 0? Yes!


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Informal Definitions of Some Types of Limits

The limit of a sequence, which is just a function ofpositive whole numbers: limn→∞ f (n) = L if (and only if)the output value f (n) stays arbitrarily close to L as long asn is sufficiently large.

Example: limn→∞

n + 1

n= 1

The limit of a function of a real variable as its inputapproaches a specified value: limx→a f (x) = L if (and onlyif) the output value f (x) stays arbitrarily close to L aslong as x is sufficiently close to a.


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Exercise: Evaluate the Limit, If It Exists

limx→0

sin x

|x |


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Exercise: Evaluate the Limit, If It Exists

1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

-10 -5 5 10

f x( ) = sin x( )

x

It does not exist. But we can consider the weaker notion of alimit as the input approaches from the left (below) or right(above). These do exist:

limx→0−

sin x

|x |= −1

limx→0+

sin x

|x |= 1


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Informal Definitions of Left and Right Limits

The limit of a function of a real variable as its inputapproaches a specified value from the left (below):limx→a− f (x) = L if (and only if) the output value f (x)stays arbitrarily close to L as long as x is sufficiently closeto a and also less than a.

The limit of a function of a real variable as its inputapproaches a specified value from the right (above):limx→a+ f (x) = L if (and only if) the output value f (x)stays arbitrarily close to L as long as x is sufficiently closeto a and also greater than a.


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Infinite Limits & Limits at Infinity

Examples:

limx→0

1

x2= +∞ (“+” is understood if you don’t specify)

limx→0

−1

x2= −∞

limx→∞

1

x= 0


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Exercises: Evaluate Each Limit (If It Exists)

1 limx→01x

2 limx→0+1x

3 limx→0−1x

4 limx→∞1x

5 limx→−∞1x

6 limx→∞1+xx

7 limx→∞√

x2

x

8 limx→−∞√

x2

x


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1 limx→01x does not exist.

2 limx→0+1x =∞

3 limx→0−1x = −∞

4 limx→∞1x

5 limx→−∞1x

6 limx→∞1+xx

7 limx→∞√

x2

x

8 limx→−∞√

x2

x


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2 limx→0+1x =∞

3 limx→0−1x = −∞

4 limx→∞1x = 0

5 limx→−∞1x = 0

6 limx→∞1+xx

7 limx→∞√

x2

x

8 limx→−∞√

x2

x


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2 limx→0+1x =∞

3 limx→0−1x = −∞

4 limx→∞1x = 0

5 limx→−∞1x = 0

6 limx→∞1+xx = 1

7 limx→∞√

x2

x

8 limx→−∞√

x2

x


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2 limx→0+1x =∞

3 limx→0−1x = −∞

4 limx→∞1x = 0

5 limx→−∞1x = 0

6 limx→∞1+xx = 1

7 limx→∞√

x2

x = 1

8 limx→−∞√

x2

x = −1


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9 limx→∞ x2

10 limx→−∞ x2

11 limx→∞ x3

12 limx→−∞ x3

13 limx→∞sin xx

14 limx→−∞sin xx

15 limx→∞ sin(

1x

)16 limx→−∞ sin

(1x

)


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9 limx→∞ x2 =∞

10 limx→−∞ x2 =∞

11 limx→∞ x3

12 limx→−∞ x3

13 limx→∞sin xx


15 limx→∞ sin(

1x


(1x

)


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9 limx→∞ x2 =∞

10 limx→−∞ x2 =∞

11 limx→∞ x3 =∞

12 limx→−∞ x3 = −∞

13 limx→∞sin xx


15 limx→∞ sin(

1x


(1x

)


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9 limx→∞ x2 =∞

10 limx→−∞ x2 =∞

11 limx→∞ x3 =∞

12 limx→−∞ x3 = −∞

13 limx→∞sin xx = 0

14 limx→−∞sin xx = 0

15 limx→∞ sin(

1x


(1x

)


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9 limx→∞ x2 =∞

10 limx→−∞ x2 =∞

11 limx→∞ x3 =∞

12 limx→−∞ x3 = −∞

13 limx→∞sin xx = 0

14 limx→−∞sin xx = 0

15 limx→∞ sin(

1x

)= 0

16 limx→−∞ sin(

1x

)= 0


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Exercise: Provide Informal Definitions

1 limx→a

f (x) = +∞ if (and only if) . . .

2 limx→a−

f (x) = +∞ if (and only if) . . .

3 limx→a+

f (x) = +∞ if (and only if) . . .

4 limx→a

f (x) = −∞ if (and only if) . . .

5 limx→a−

f (x) = −∞ if (and only if) . . .

6 limx→a+

f (x) = −∞ if (and only if) . . .

7 limx→∞

f (x) = L if (and only if) . . .

8 limx→−∞


9 limx→∞

f (x) =∞ if (and only if) . . .

10 limx→−∞


11 limx→∞

f (x) = −∞ if (and only if) . . .

12 limx→−∞

f (x) = −∞ if (and only if) . . .


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Formal & Precise Definition:Finite Limit at a Finite Value

Definition. limx→a

f (x) = L if (and only if), given any

positive real number ε, there is a positive real number δsuch that

0 < |x − a| < δ ⇒ |f (x)− L| < ε.

Remark: The condition that 0 < |x − a| < δ means thatthe value of x is within δ of a, but not equal to a. Thelimit at a requires nothing when the value of x is equal toa, where the value of f (x) may be undefined.

Remark: The consequence that |f (x)− L| < ε means thatthe value of f (x) is within ε of the limiting value L. It isdoes not matter whether or not f (x) is equal to L forsome values of x satisfying the condition, hence there isno requirement that 0 < |f (x)− L|.


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Illustrative Contrasting Examples

For example, if the function f is a constant functiondefined by f (x) = c , c ∈ R, and if a is any real number,limx→a f (x) = c because f (x) = c , and hence|f (x)− c | = |c − c | = 0 < ε, for every value of x .

On the other hand, if the function g is defined byg(x) = x2

x , then limx→0 g(x) = 0, even though g(x) is notequal to 0 for any value of x . Note that g(x) is notdefined for x = 0; 0 is not in the domain of g .

These examples illustrate the importance of attention todetails in a precise definition.


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Formal & Precise Definition:Finite Limits from the Left & Right

Definition. limx→a−

f (x) = L if (and only if), given any

positive real number ε, there is a positive real number δsuch that

a− δ < x < a⇒ |f (x)− L| < ε.

Exercise:

Provide a precise, formal definition: limx→a+

f (x) = L if (and

only if) . . ..


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Formal & Precise Definition: Infinite Limits

Just as f (x) being arbitrarily close to L means that|f (x)− L| < ε, where ε may be any positive real number(this is the arbitrary part), f (x) being arbitrarily close to+∞ means that f (x) > N, where N may be any positiveinteger.

We think of ε as getting smaller and smaller; we think ofN as getting larger and larger. (We could have let N beany real number, but it may as well be an integer.)

Thus we make the definition of a positive infinite limitprecise and formal as follows:Definition. lim

x→af (x) = +∞ if (and only if), given any

positive integer N, there is a positive real number δ suchthat

0 < |x − a| < δ ⇒ f (x) > N.


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Exercises: Provide Precise & Formal Definitions

1 Definition. limx→a−

f (x) = +∞ if (and only if) . . ..

2 Definition. limx→a+

f (x) = +∞ if (and only if) . . ..

3 Definition. limx→a

f (x) = −∞ if (and only if) . . ..

4 Definition. limx→a−

f (x) = −∞ if (and only if) . . ..

5 Definition. limx→a+

f (x) = −∞ if (and only if) . . ..

6 Definition. limx→∞


7 Definition. limx→−∞







f (x) = −∞ if (and only if) . . .


f (x) = −∞ if (and only if) . . .


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Theorem:The Limit of a Sum is the Sum of the Limits

Theorem

If limx→a f (x) and limx→a g(x) exist and are finite, thenlimx→a f (x) + g(x) = limx→a f (x) + limx→a g(x).

Applying the Theorem:

Example: limx→0

sin x

x+ x2 = 1 + 0 = 1.

Example: The theorem does not apply to limx→0sin xx + 1

x ,since limx→0

1x does not exist.

Example: The theorem does not apply to limx→0sin xx + 1

x2 ,since limx→0

1x2 =∞ is not finite.


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How to Show a Theorem is True

The consequence of the theorem need only hold forinstances that satisfy the condition.

If the condition is false, there is nothing to show!

Therefore, to show that the theorem is true, we assumethe condition is true; under this assumption, we mustlogically demonstrate the truth of the consequence.

Please note that this assumption is provisional; thecondition is certainly not true in all instances!

Please also note that we must take care to assume nothingbeyond the stated condition.


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Restating the Theorem Often Helps

Some labels make both the condition and the consequenceeasier to state and work with:

Let limx→a f (x) = L.Let limx→a g(x) = M.

Substituting these labels, we have the followingrestatement of the theorem:

Theorem

If limx→a f (x) = L and limx→a g(x) = M, thenlimx→a f (x) + g(x) = L + M.


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Using Definitions to Work with the Condition

The condition that limx→a f (x) = L means that we canmake |f (x)− L| as small as we like, as long as x issufficiently close to a; sufficiently close means0 < |x − a| < δ, for a suitable positive real number of δ.

Similarly, we can make |g(x)−M| as small as we like.

Key point: for the smaller value of δ, both |f (x)− L| and|g(x)−M| will be as small as we like.

So . . . how small do we need them to be?


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Using Definitions to Work with the Consequence

The consequence that limx→a f (x) + g(x) = L + Mmeans, given any positive real number ε, there is apositive real number δ such that 0 < |x − a| < δ issufficient to ensure that |f (x) + g(x)− (L + M)| < ε (thatis, 0 < |x − a| < δ ⇒ |f (x) + g(x)− (L + M)| < ε).

To show this is true, we must consider an arbitrary positivereal number ε and show that a suitable δ exists for that ε.

We will find a suitable δ by making |f (x)− L| and|g(x)−M| small enough to ensure that|f (x) + g(x)− (L + M)| < ε.


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A Picture Shows Why the Theorem is True

} }} }f(x) g(x)

L |f(x)-L| |g(x)-M|M

} }

How small must |f (x)− L| and |g(x)−M| be to ensurethat |f (x) + g(x)− (L + M)| < ε?


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Conclusion of the Proof!

} }} }f(x) g(x)

L < e/2 < e/2M

} }

If |f (x)− L| < ε2 and |g(x)−M| < ε

2 , then|f (x) + g(x)− (L + M)| = |f (x)− L + g(x)−M| ≤|f (x)− L|+ |g(x)−M| < ε

2 + ε2 = ε.


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Theorem:The Limit of a Product is the Product of the Limits

Theorem

If limx→a f (x) and limx→a g(x) exist and are finite, thenlimx→a f (x)g(x) = limx→a f (x) · limx→a g(x).

Applying the Theorem:

Example: limx→0

(sin x

x

)(x2 + 2x

x

)= (1)(2) = 2.

Letting limx→a f (x) = L and limx→a g(x) = M, we againhave a restatement in a form that is easier to prove:

Theorem

If limx→a f (x) = L and limx→a g(x) = M, thenlimx→a f (x)g(x) = LM.


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Using Definitions to With the Condition and theConsequence

Again, our condition tells us that we can make |f (x)− L|and |g(x)−M| as small as we like, as long as0 < |x − a| < δ, for a suitable positive real number of δ ineach case.

Again, a simple but key observation is that a single choiceof δ will work in both cases.

Again, we must consider an arbitrary positive real numberε. To prove the current theorem, we must show that asuitable δ exists to ensure that |f (x)g(x)− LM| < ε..

Again will find a suitable δ by making |f (x)− L| and|g(x)−M| small enough to ensure that|f (x)g(x)− LM| < ε.

So . . . how small do we need them to be?


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A Picture Shows Why the Theorem is True

}}} }L |f(x)-L|

M|f(x)-L|

|g(x)-M| |g(x)-M||f(x)-L||g(x)-M|L

M

}

}

f(x)

g(x)ML

The picture shows that |f (x)g(x)− LM| ≤|g(x)−M||L|+ |g(x)−M||f (x)− L|+ |M||f (x)− L|.


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The Conclusion of the Proof!

}}} }L

< e/3M & < L

< M(e/3M)= e/3

< e/3L< (e/3L)L = e/3 < (e/3L)L = e/3

M

}

}

f(x)

g(x)ML

Choose δ such that 0 < |x − a| < δ ⇒:

|g(x)−M| < ε3|L| , and

|f (x)− L| < |L|, and|f (x)− L| < ε

3|M| .

Then |f (x)g(x)− LM| < ε3|L| · |L|+

ε3|L| · |L|+ |M| ·

ε3|M| =

ε3 + ε

3 + ε3 = ε.


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Limits of Differences & Quotients

Analogous theorems hold for subtraction and division:

The limit of a difference is the difference of the limits.The limit of a quotient is the quotient of the limits, aslong as the limit in the denominator is not zero.

Once we know the necessary limits exist, we can readilydeduce the correct formulas by “working backwards”.

Let limx→a f (x) = L, limx→a g(x) = M,

limx→a f (x)− g(x) = N, and limx→af (x)g(x) = O.

By the sum theorem,N + M = limx→a f (x)− g(x) + limx→a g(x) =limx→a

(f (x)− g(x)

)+ g(x) = limx→a f (x) = L.

Thus, since N + M = L, it must be that N = L−M.Exercise: Show that as long as M 6= 0, O = L

M .


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Complicated Limits from Simpler Ones

The results about limits of sums, products, differences,and quotients provide rules, or laws, for evaluating limits.

Used in combination with each other and with otherresults about limits, they provide more general rules.

The purpose of these rules is to evaluate complicatedlimits using our knowledge of simpler ones.

However, it is important to recognize that these rules arenot the only means of evaluating limits.

For example:

limx→01x does not exist, but limx→0 x2 · 1

x = 0.limx→1 x − 1 = 0, so the quotient rule does not apply, but

limx→1x2−1x−1 is defined.

Exercise: What is it? Why?


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Limits of Positive Integer Powers

By repeated application of the product rule, we get thepositive integer power rule:

limx→a[f (x)]2 = limx→a f (x) · f (x) =limx→a f (x) · limx→a f (x) = [limx→a f (x)]2.limx→a[f (x)]3 = limx→a[f (x)]2 · f (x) =[limx→a f (x)]2 · limx→a f (x) = [limx→a f (x)]3.. . .limx→a[f (x)]n = [limx→a f (x)]n, for any positive integer n.


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Limits of Negative Integer Powers

Combining the rule for positive powers with an applicationof the quotient rule, we obtain a similar formula fornegative integer exponents:

limx→a

[f (x)]−n = limx→a

1

[f (x)]n

=1

limx→a[f (x)]n=

1

[limx→a f (x)]n= [ lim

x→af (x)]−n.

(Here we used our knowledge, based directly on thedefinition of the limit, that limx→a 1 = 1.)


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Limits of Fractional Powers

Once we know that the necessary limits exist, we cangeneralize the formula to fractional powers using thestandard strategy for working with inverse operations:

Let limx→a f (x) = L and let limx→a[f (x)]1n = M

Mn = [limx→a[f (x)]1n ]n = limx→a

([f (x)]

1n

)n=

limx→a f (x) = L.

Since Mn = L, M = L1n .

Applying the power rule, we get:limx→a[f (x)]

mn = limx→a

([f (x)]

1n

)m=(

limx→a[f (x)]1n

)m=([limx→a f (x)]

1n

)m= [limx→a f (x)]

mn .

Remember that fractional powers are not always defined.

(Ex. (−2)12 is not defined.)

This rule also works for irrational powers (when defined).

The limit of a power is the power of the limits!


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Algebraic Combinations

A rational combination of functions is a function obtainedby combining the outputs of those functions using thebasic operations + and · and their inverse operations −and /.Examples:

Given functions f and g , the function f + g is defined byf + g(x) = f (x) + g(x).Given functions f and g , the function fg is defined byfg(x) = f (x)g(x).

Given functions f , g , and h, the function f−g2

h is defined

by f−g2

h (x) = f (x)−[g(x)]2

h(x) .

More generally, if roots (that is, fractional exponents) arealso involved, we will call the result an algebraiccombination.


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Summary:Limits of Algebraic Combinations

The theorems we have just developed may be succinctlysummarized as follows:

The Limitof an Algebraic Combination

is the Combination of the Limits(If The Combination of the Limits is Defined).

This result and the other limit theorems have analoguesfor one-sided limits, infinite limits, and limits at infinity;the necessary modifications and restrictions are generallyclear, and we will not discuss them separately.


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Examples

1 Suppose limx→a f (x) = 2, limx→a g(x) = 3, andlimx→a h(x) = 5. Compute:

limx→a

f (x) + g(x)

f (x)h(x)

2 Suppose limx→a f (x) = 2, limx→a+ g(x) = 3, andlimx→a+ h(x) = 5. Compute:

limx→a+

f (x) + g(x)

f (x)h(x)


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Examples

1 Suppose limx→a f (x) = 2, limx→a g(x) = 3, andlimx→a h(x) = 5. Compute:

limx→a

f (x) + g(x)

f (x)h(x)

Solution: limx→af (x)+g(x)f (x)h(x) = 2+3

(2)(5) = 12

2 Suppose limx→a f (x) = 2, limx→a+ g(x) = 3, andlimx→a+ h(x) = 5. Compute:

limx→a+

f (x) + g(x)

f (x)h(x)

Solution: Since one-sided limits exist and agree with thetwo-sided limit if it exists, we get the same answer.


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The Limit of a Constant Function

Theorem

If c is any real number, limx→a

c = c.

This rather obvious fact follows directly from the definition oflimit:

Proof.

Given any positive real number ε, let δ = 1, or anything elseyou like! 0 < |x − a| < 1⇒ |c − c| = 0 < ε.


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The Limit of the Identity Function

Theorem

limx→a

x = a.

This rather obvious fact also follows directly from the definitionof limit:

Proof.

Given any positive real number ε, let δ = ε.0 < |x − a| < ε⇒ |x − a| < ε.


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It follows from the the theorems we have just developed that iff is defined as an algebraic combination of constant functionsand the identity function i(x) = x , and if f (a) is defined, thenlimx→a f (x) = f (a). Here are some examples:

1 limx→0

x2 + 5x + 3

2 limx→2

x2 + 5x + 3

3 limx→2

3x3 + 2x2 +√

5x + π

4 limx→0

x2 + 3

x − 1

5 limx→8

x23 + 2x

43 + 1


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1 limx→0

x2 + 5x + 3 = 02 + 5(0) + 3 = 3.

2 limx→2

x2 + 5x + 3

3 limx→2

3x3 + 2x2 +√

5x + π

4 limx→0

x2 + 3

x − 1

5 limx→8

x23 + 2x

43 + 1


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1 limx→0

x2 + 5x + 3 = 02 + 5(0) + 3 = 3.

2 limx→2

x2 + 5x + 3 = 22 + 5(2) + 3 = 4 + 10 + 3 = 17.

3 limx→2

3x3 + 2x2 +√

5x + π

4 limx→0

x2 + 3

x − 1

5 limx→8

x23 + 2x

43 + 1


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1 limx→0

x2 + 5x + 3 = 02 + 5(0) + 3 = 3.

2 limx→2

x2 + 5x + 3 = 22 + 5(2) + 3 = 4 + 10 + 3 = 17.

3 limx→2

3x3 + 2x2 +√

5x + π = 3(23) + 2(22) +√

5 · 2 + π =

24 + 8 + 2√

5 + π = 32 + 2√

5 + π.

4 limx→0

x2 + 3

x − 1

5 limx→8

x23 + 2x

43 + 1


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1 limx→0

x2 + 5x + 3 = 02 + 5(0) + 3 = 3.

2 limx→2

x2 + 5x + 3 = 22 + 5(2) + 3 = 4 + 10 + 3 = 17.

3 limx→2

3x3 + 2x2 +√

5x + π = 3(23) + 2(22) +√

5 · 2 + π =

24 + 8 + 2√

5 + π = 32 + 2√

5 + π.

4 limx→0

x2 + 3

x − 1=

3

−1= −3.

5 limx→8

x23 + 2x

43 + 1


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1 limx→0

x2 + 5x + 3 = 02 + 5(0) + 3 = 3.

2 limx→2

x2 + 5x + 3 = 22 + 5(2) + 3 = 4 + 10 + 3 = 17.

3 limx→2

3x3 + 2x2 +√

5x + π = 3(23) + 2(22) +√

5 · 2 + π =

24 + 8 + 2√

5 + π = 32 + 2√

5 + π.

4 limx→0

x2 + 3

x − 1=

3

−1= −3.

5 limx→8

x23 + 2x

43 + 1 = 8

23 + 2(8

43 ) + 1 = 22 + 2(24) + 1 =

4 + 2(16) + 1 = 37.


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The System of Functions

Real-valued functions are elements of a system withoperations, just like numbers. Sums, products, differences,and quotients of functions are defined by applying theseoperations to their outputs.

Just as there are important subsystems of the real numbersystem, such as the integers, there are importantsubsystems of the function system.

There are strong analogies between the number systemand the function system.

Although not perfect, these analogies help us see whatthese different systems have in common.

Seeing the relationships between different systems helps usorganize our knowledge.


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Polynomial Functions are Analogous to the Integers

The most special and basic real numbers are . . .

The most special and basic types of functions are . . .


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Polynomial Functions are Analogous to Integers

The most special and basic real numbers are 0 and 1.

The most special and basic types of functions are . . .


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The most special and basic types of functions are constantfunctions (that is, functions of the form f (x) = c) and theidentity function (i(x) = x). Note that these are thepolynomials of degree 0 and 1.


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The most special and basic types of functions are constantfunctions and the identity function.

Combinations of 0 and 1 using addition and subtractiongive us the integers. Multiplication of integers is justrepeated addition, so no new numbers are generated if weinclude multiplication, too.

Combinations of constant functions and the identityfunction using addition, subtraction, and multiplicationgive us the polynomial functions.

Example: P(x) = 2x7 + x3 − 3x2 + 5.


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Rational Functions are Analogous toRational Numbers

If we include division in our combinations of numbers(starting with 0 and 1), we get the rational numbers.

If we include division in our combinations of functions(starting with constant functions and the identityfunction), we get the rational functions.

In other words, a rational function is a rationalcombination of constant functions and the identityfunction.

Examples:

Q(x) = x2 + x−3

R(x) =x2 + x−3

x − x−1

S(x) =2x7 + x3 − 3x2 + 5

x2 − x + 3


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A Rational Function is a Quotient of Polynomials

Just as any rational number may be simplified to aquotient of two integers, any rational function may besimplified the quotient of two polynomials, as in theexample S(x) = 2x7+x3−3x2+5

x2−x+3

Exercise: Do this for the two examples below.

1 Q(x) = x2 + x−3

2 R(x) =x2 + x−3

x − x−1


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x2−x+3


1 Q(x) = x2 + x−3 = x2 +1

x3=

x5 + 1

x3.

2 R(x) =x2 + x−3

x − x−1


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x2−x+3


1 Q(x) = x2 + x−3 = x2 +1

x3=

x5 + 1

x3.

2 R(x) =x2 + x−3

x − x−1=

x5+1x3

x2−1x

=x5 + 1

x3· x

x2 − 1=

x5 + 1

x4 − x2.


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Algebraic Numbers

A number that is a root of a polynomial with integercoefficients is called an algebraic number. (No newnumbers are generated if we include rational coefficients,since we can find a common denominator and multiply toget an equivalent equation.)

Examples:√

5 is an algebraic number, since it is a root of x2 − 5(that is, a solution to the equation x2 − 5 = 0).1+√

52 is an algebraic number, since it is the solution to the

equation x2 − x − 1 = 0.Any algebraic combination of algebraic numbers, such as

3√

1 +√

2, is an algebraic number.Remark: Not all algebraic numbers can be obtained byrepeatedly taking and combining radicals. This deep resultdue to Galois lies at the genesis of modern algebra.


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Algebraic Functions are Analogous toAlgebraic Numbers

In analogous fashion, functions such as the one defined byg(x) =

√x2 − x − 1 are called algebraic functions.

Any algebraic combination of algebraic functions is analgebraic function.

To find limits of algebraic functions, we use our theoremon the limits of algebraic combinations. (We can “plug in”the limiting input value, as long as the correspondingoutput value is defined.)

Note that the domain of an algebraic function that is nota polynomial may be restricted.

1 Exercise: What is the domain of the function f defined by

f (x) = x5+1x4−x2 ?

2 Exercise: What is the domain of the function g defined byg(x) =

√x2 − x − 1?


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Finding the Domain of a Function

1 For the function f defined by f (x) = x5+1x4−x2 , the domain of

f is the set of real numbers other than 0,−1, or 1:

Domain(f ) = R \ {0,−1, 1}.Reasoning: f (x) is defined except where the denominatorx4 − x2 is zero. x4 − x2 = x2(x2 − 1) = x2(x + 1)(x − 1),thus x4 − x2 = 0⇔ x = 0 or x = −1 or x = 1.

2 For the function g defined by g(x) =√

x2 − x − 1, thedomain of g is the set of real numbers that are less than1−√

52 or greater than 1+

√5

2 .

Domain(g) = (−∞, 1−√

52 ) ∪ ( 1+

√5

2 ,∞).

Reasoning: g(x) is defined except where x2 − x − 1 < 0.

Since the roots of x2 − x − 1 are 1±√

52 , it factors as

x2 − x − 1 = (x − 1−√

52 )(x − 1+

√5

2 ). The value ofx2 − x − 1 is negative between the two roots.


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Transcendental Functions are Analogous toTranscendental Numbers

Any numbers that do not fit the classification justoutlined, but rather transcend it (lie beyond its reach) arecalled transcendental numbers.

Most real numbers (in a sense of most that can beprecisely defined) are transcendental, but few of themhave names. You are familiar with one or two of them:

π is transcendental.e, the base of the natural logarithm, is transcendental.

In analogous fashion, functions that are not algebraic arecalled transcendental functions. Most importantly:

The trigonometric functions and their inverse functions.Exponential functions, meaning those where the exponentvaries, as in f (x) = 2x , and their inverse (logarithmic)functions. These will be studied thoroughly in Calculus 2.


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Summary: Structure of the Number System

The set of real numbers divides into two disjoint subsets,the algebraic numbers and the transcendental numbers.

The algebraic numbers are roots of polynomials withinteger (equivalently, rational) coefficients.The transcendental numbers are not.

The algebraic numbers includes, as a subset, the set ofrational numbers, which includes, as a subset, the set ofintegers. Rational numbers are the roots of linearpolynomials (for example, 2

3 is the solution to 3x − 2 = 0);integers are the roots of monic linear polynomials, that is,those with leading coefficient 1 (for example, 3 is thesolution to x − 3 = 0).

Examples:√

2 is algebraic, but not rational; 23 is rational,

but not an integer; 3 is an integer; π is transcendental.


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Summary: Structure of the Function System

The set of real-valued functions divides into two disjointsubsets, algebraic functions and transcendental functions.

The algebraic functions are roots of polynomial functionalequations with polynomial coefficients (for example,(x2 + x)[f (x)]3 − x3[f (x)]2 + (x5 − 2x + 1) = 0, whereequality must hold for all input values of x where f can bedefined).The transcendental functions are not.In practice, the algebraic functions we can easily writedown are combinations of constant functions and theidentity function using the elementary operations(+,−, ·,÷) and roots, just as the algebraic numbers wecan easily write are combinations of 0 and 1 using theelementary operations and roots.


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Summary: Structure of the Function System(Continued)

The set of algebraic functions includes, as a subset, the setof rational functions, which includes, as a subset, the setof polynomial functions. Rational functions are the rootsof linear functional equations (for example, x2−1

x+2 is the

solution to (x + 2)f (x)− (x2 − 1) = 0); polynomials arethe roots of monic linear functional equations, (forexample, x2 − 1 is the solution to f (x)− (x2 − 1) = 0).

Examples: the function defined by f (x) =√

x2 − 1 isalgebraic, but not rational; the function defined byf (x) = x2−1

x+2 is rational, but not polynomial; the function

defined by f (x) = x2 − 1 is polynomial; sine istranscendental.


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Comparing Functions

Theorem

Suppose f (x) ≤ g(x) in a neighborhood surrounding (butexcluding) x = a (that is, in an open interval on either side ofa) and the limits limx→a f (x) and limx→a g(x) exist. Thenlimx→a f (x) ≤ limx→a g(x).

Proof.

Let L = limx→a f (x) and M = limx→a g(x); we wish to showL ≤ M. By way of contradiction, if it were the case thatL > M, we could consider ε = L−M

2 . For x sufficiently close toa (that is, 0 < |x − a| < δ for suitably small δ), we would havef (x) > L− ε = M+L

2 and g(x) < M + ε = M+L2 ; hence,

f (x) > g(x) when 0 < |x − a| < δ, strongly contradicting ourhypothesis that f (x) ≤ g(x) in a neighborhood of x = a!


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Getting the Strongest Conclusion

Examining our argument, we see that the condition thatf (x) ≤ g(x) in a neighborhood surrounding (butexcluding) x = a actually implies something stronger:

If limx→a g(x) exists, then for any positive real number ε,there is a positive real number δ such that0 < |x − a| < δ ⇒ f (x) < limx→a g(x) + ε, whether or notlimx→a f (x) itself exists.Similarly, if limx→a f (x) exists, then for any positive realnumber ε, there is a positive real number δ such that0 < |x − a| < δ ⇒ g(x) > limx→a f (x)− ε, whether or notlimx→a g(x) itself exists.

These observations lead immediately to the following veryimportant theorem.


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The Squeeze Theorem

Theorem (Squeeze Theorem)

If f (x) ≤ g(x) ≤ h(x) in a neighborhood surrounding (butexcluding) x = a and limx→a f (x) = limx→a h(x) = L, thenlimx→a g(x) = L.

Proof.

For any positive real number ε, there is a positive real numberδ such that 0 < |x − a| < δ ⇒ g(x) < L + ε (since g(x) < h(x)in a neighborhood surrounding (but excluding) x = a, andlimx→a h(x) = L) and g(x) > L− ε (since f (x) < g(x) in aneighborhood surrounding (but excluding) x = a, andlimx→a f (x) = L).


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Exercise: Using the Squeeze Theorem

The Squeeze Theorem is often used to evaluate limits oftranscendental functions by comparing them to algebraicfunctions whose limits we know.

Exercise: Evaluate limx→0

x3 sin

(1

x

).

Explain and justify your calculation.

There is another very instructive example in the text!


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Continuous Functions

Definition

A function f is continuous at a if limx→a f (x) = f (a).

Exercises:

Define continuous from the left at a.Define continuous from the right at a.

Important observation: if f is continuous at a, therequirement that x 6= a may be dropped from thecondition of the limit definition, since f (a) is defined andequal to limx→a f (x): for any positive real number ε, thereis a positive real number δ such that|x − a| < δ ⇒ |f (x)− f (a)| < ε.


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Continuity

Algebraic Functions are Continuouson their Domains

If f is continuous at every point of an interval I , we say fis continuous on I ; if f is defined on only one side of anendpoint of the interval, continuity from the appropriateside is understood.

If f is continuous at every point in its domain, we say f iscontinuous on its domain. If f is continuous at every realnumber, we simply say f is continuous.

It follows from the theorem on limits of algebraiccombinations that every algebraic function is continuouson its domain. In particular, any polynomial is continuous.

The trigonometric functions are also continuous on theirdomains. In particular, sine and cosine are continuous.

Any algebraic combination of continuous functions iscontinuous.


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Exercise: Which of the Following Functions areContinuous at 1?

1 The function f defined by f (x) = x + 1.

2 The function g defined by g(x) = x2 − 1.

3 The function h defined by h(x) = 1x−1 .

4 The function j defined by j(x) = x2−1x−1 .

5 The function k defined by k(x) = 1x .


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1 The function f defined by f (x) = x + 1. Yes, f iscontinuous at 1.






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2 The function g defined by g(x) = x2 − 1. Yes, g iscontinuous at 1.





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3 The function h defined by h(x) = 1x−1 . No, h is not

continuous at 1.




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continuous at 1.

4 The function j defined by j(x) = x2−1x−1 . No, j is not

continuous at 1, even though limx→1 j(x) = 2, since j(1)is not defined.



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continuous at 1.

4 The function j defined by j(x) = x2−1x−1 . No, j is not

continuous at 1, even though limx→1 j(x) = 2, since j(1)is not defined.

5 The function k defined by k(x) = 1x . Yes, k is continuous

at 1.


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Exercise: Which of the Following Functions areContinuous on (0,∞)?

1 The function f defined by f (x) = x + 1.






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1 The function f defined by f (x) = x + 1. Yes.






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2 The function g defined by g(x) = x2 − 1. Yes.





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3 The function h defined by h(x) = 1x−1 . No.




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4 The function j defined by j(x) = x2−1x−1 . No.



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4 The function j defined by j(x) = x2−1x−1 . No.

5 The function k defined by k(x) = 1x . Yes.


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Graphs of Continuous Functions

Continuity is reflected in the graph of the function. If afunction is continuous on an interval, its graph will haveno gaps or breaks on that interval.

The function f defined by f (x) = (x − 1)(x − 2)(x − 3) iscontinuous.

10

8

6

4

2

-2

-4

-6

-8

-10

-12

-3 -2 -1 1 2 3 4 5 6

The function g defined by g(x) = 1x is continuous on

(−∞, 0) and (0,∞), but not at 0.8

6

4

2

-2

-4

-6

-8

-10 -5 5 10


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Removable Discontinuities

As we know, function f is discontinuous at a if it is notthe case that limx→a f (x) = f (a). There are three waysthis can happen:

limx→a f (x) does not exist.limx→a f (x) = f (a) exists, but f (a) is not defined.limx→a f (x) = f (a) exists and f (a) is defined, but thesenumbers are not equal.

In the latter two cases, when limx→a f (x) = f (a) exists,we can remove the discontinuity by defining, or redefining,f (a) = limx→a f (x).

In these cases, the discontinuity is said to be removable.


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Example

The function f defined by f (x) = x sin 1x is not defined for

x = 0; thus, f is not continuous at x = 0.

However, as we know, limx→0 f (x) = 0. Defining f (0) = 0removes the discontinuity, creating the continuous function

f (x) =

{x sin 1

x , if x 6= 0

0 , if x = 0


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Computing Limits by Removing Discontinuities

Suppose f (a) is not defined but there is a function g thatis continuous at a such that f (x) = g(x) for all values ofx except a.

Then limx→a f (x) = g(a). Replacing f with g removes thediscontinuity of f at a.

Often we can recognize a formula for such a function g byalgebraic manipulation.

Example:x2 − 1

x − 1=

(x + 1)(x − 1)

x − 1= x + 1 for all values

of x except 1. Thus, limx→1

x2 − 1

x − 1= 1 + 1 = 2.


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Examples with One-sided Continuityat Some Points

One-sided continuity arises in applications.

For example, suppose a fair cubical die with sides labeled1, 2, 3, 4, 5, and 6 is tossed. Let F (x) be the probabilitythat the number on the top face is less than or equal to x .

What does the graph of F look like?

Where is this function continuous?

At the points where it is not continuous, is it continuousfrom one side? Which side?


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The Graph of F

1

5/6

2/3

3/6

1/3

1/6

1 2 3 4 5 6

y=F(x)

F is continuous except at 1, 2, 3, 4, 5, and 6, where it iscontinuous from the right but not the left.

Exercise: Describe a function that is continuous except at1, 2, 3, 4, 5, and 6, where it is continuous from the leftbut not the right.


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Composition of Functions

Recall that if x is in the domain of f and f (x) is in thedomain of g , then the function f ◦ g , called thecomposition of f and g , is defined by g ◦ f (x) = g(f (x)).

Composition is another operation on functions, giving thesystem of functions additional structure.

Exercise: What is the domain of g ◦ f in terms of thedomains of f and g?

Compositions often arise in applications. For example, ifthe volume of water in a tank is a function of time,v = f (t), and the depth of the water in the tank is afunction of its volume, d = g(v), then the depth of thewater at each time is given by the composition of thesetwo functions, d = g(f (t)) = g ◦ f (t).


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Composition of Functions

Recall that if x is in the domain of f and f (x) is in thedomain of g , then the function f ◦ g , called thecomposition of f and g , is defined by g ◦ f (x) = g(f (x)).

Composition is another operation on functions, giving thesystem of functions additional structure.

Domain(g ◦ f ) = {x ∈ domain(f ) : f (x) ∈ domain(g)} =f −1(domain(g)), the preimage under f of domain(g).

Compositions often arise in applications. For example, ifthe volume of water in a tank is a function of time,v = f (t), and the depth of the water in the tank is afunction of its volume, d = g(v), then the depth of thewater at each time is given by the composition of thesetwo functions, d = g(f (t)) = g ◦ f (t).


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Limits of CompositionsInvolving Continuous Functions

Theorem

If limx→a f (x) = b and g is continuous at b, thenlimx→a g(f (x)) = g(b).

Proof.

(Informal.) For x sufficiently close (but not equal) to a, f (x) issufficiently close to b so that g(f (x)) is as close as we like tog(b); recall that we need not worry about the possibility thatf (x) = b. (We use here that limx→b g(x) = g(b).)

Corollary

If f is continuous at a and g is continuous at b = f (a), theng ◦ f is continuous at a. In particular, the composition ofcontinuous functions is continuous.


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Exercises: Evaluate the Limits;Explain and Justify Your Solutions

1 limx→0

sin(

x2 − 3x +π

2

)2 lim

x→π2

cos

(x2 − π2

4

x − π2

)

3 limx→a

x2 − a2

x − a(Answer in terms of a.)

4 limh→0

1x+h −

1x

h(Answer in terms of x . Assume x 6= 0.)

5 limh→0

√x + h −

√x

h. (Answer in terms of x . Assume x > 0.

For x = 0, only the limit from above exists, since x + hwould be negative when h is negative, even if h is very

small. Hint: Multiply by√

x+h+√

x√x+h+

√x

.)


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1 limx→0

sin(

x2 − 3x +π

2

)= sin

(π2

)= 1, since the sine

function and the polynomial function defined by

f (x) = x2 − 3x +π

2are both continuous.


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2 limx→π

2

cos

(x2 − π2

4

x − π2

)= cos(π) = −1, since the cosine

function is continuous, and limx→π

2

x2 − π2

4

x − π2

=

limx→π

2

(x + π2 )(x − π

2 )

x − π2

= limx→π

2

x +π

2=π

2+π

2= π.

Substitutingπ

2for x to compute lim

x→π2

x +π

2is justified

because the polynomial function defined by f (x) = x +π

2is continuous and removes the discontinuity at

π

2.


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3 limx→a

x2 − a2

x − a= lim

x→a

(x + a)(x − a)

x − a= lim

x→ax + a

= a + a = 2a.

Substituting a for x to compute limx→a

x + a is justified

because the polynomial function defined by f (x) = x + ais continuous and removes the discontinuity at a.


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4 limh→0

1x+h −

1x

h= lim

h→0

x−(x+h)x(x+h)

h= lim

h→0

−h

x(x + h)· 1

h=

limh→0

−1

x(x + h)=−1

x2, as long as x 6= 0.

Substituting 0 for h to compute limh→0

−1

x(x + h)is justified

because the rational function−1

x(x + h)is continuous at

h = 0 and removes the discontinuity as long as x 6= 0.


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5 limh→0

√x + h −

√x

h= lim

h→0

√x + h −

√x

h·√

x + h +√

x√x + h +

√x

=

limh→0

x + h − x

h(√

x + h +√

x)= lim

h→0

h

h(√

x + h +√

x)=

limh→0

1√x + h +

√x

=1√

x +√

x=

1

2√

x.

Substituting 0 for h to compute limh→01√

x+h+√

xis

justified because the algebraic function1√

x + h +√

xis

continuous at h = 0 and removes the discontinuity as longas x > 0.


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Exercise: Continuity and Removable Discontinuities

At what values are the functions defined by the followingformulas continuous? For any points of discontinuity, which areremovable discontinuities? Explain and justify your answers!

1 f (x) = sin(

1x

)2 f (x) = x2 sin

(1x

)3 f (x) = sin x

x

4 f (x) = cos(x2)

5 f (x) = sin(cos(x))

6 f (x) = sin(tan(x))

7 f (x) = tan(sin(x))

8 f (x) = cos(x sin

(1x

) )


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1 f (x) = sin(

1x

)is continuous at all input values except

x = 0. The function g defined by g(x) = 1x is rational and

therefore continuous on its domain; 0 is the only point notin its domain. Sine is continuous, so the compositionsin ◦g is continuous at all points except 0. Thediscontinuity is not removable, since limx→0 sin

(1x

)does

not exist.

2 f (x) = x2 sin(

1x

)is continuous at all input values except

x = 0. Here, in addition to the arguments used above, weuse the fact that the polynomial function defined byh(x) = x2 is continuous and that the product of functionscontinuous at a point is continuous at that point. Thediscontinuity at 0 is removable, sincelimx→0 x2 sin

(1x

)= 0; defining f (0) = 0 makes f

continuous.


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3 f (x) = sin xx is continuous at all input values except 0;

since f is a quotient of continuous functions, it iscontinuous wherever it is defined. The discontinuity at 0 isremovable by defining f (0) = 1, since limx→0

sin xx = 1.


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4 f (x) = cos(x2) is continuous, since it is the compositionof continuous functions.

5 f (x) = sin(cos(x)) is continuous, since it is thecomposition of continuous functions.


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6 f (x) = sin(tan(x)) is continuous except where the tangentfunction is discontinuous, at the points of the formπ2 + kπ, where k is any integer. These discontinuities arenot removable, since tan x → ±∞ is x approaches each ofthese values from the left or right, respectively, and sineoscillates as its input value approaches ±∞.

7 f (x) = tan(sin(x)) is continuous, since sin(x) is neverequal to π

2 + kπ for any integer k. (−1 ≤ sin(x) ≤ 1.)


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8 f (x) = cos(x sin

(1x

) )is continuous except at x = 0,

since sine, cosine, and the identity function arecontinuous. The discontinuity is removable by definingf (0) = 1, since limx→0 x sin

(1x

) )= 0 and cos(0) = 1.


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Continuity of the Real Number System

Our last result about continuous functions, the importantIntermediate Value Theorem, requires the continuity (in aslightly different sense) of the real number system itself.

All of the properties of the real number system withrespect to operations and order are true for the system ofrational numbers as well.

What makes the real numbers different is the absence ofany gaps, as described by the following continuity property:

Given any two subsets of real numbers, A and B, suchthat every element of A is less than every element of B,there is a real number s that separates A from B in thesense that every element of A is less than or equal to s,whereas every element of B is greater than or equal to s.


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Continuity of the Real Number System (Continued)

For example, let A = {x : x > 0 and x2 < 2} andB = {x : x > 0 and x2 > 2}. If we restrict our attentionto the rational numbers, there is no number that separatesA from B. For any rational number r , either r 2 > 2, inwhich case a slightly smaller number b also has b2 > 2, orr 2 < 2, in which case a slightly larger number a also hasa2 < 2.

In either case, r does not separate A from B; only√

2,which is real but not rational, separates A from B.

Unlike the rationals, the real numbers vary continuously.

A valuable side benefit is that many equations with norational solutions, such as x2 − 2 = 0, have real solutions.

Nonetheless, not all equations have real solutions. (Forexample, x2 + 2 = 0 has none.)


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The Intermediate Value Theorem (IVT)

Theorem (Intermediate Value Theorem (IVT))

If f is continuous on [a, b], f (a) 6= f (b), and u is any numberbetween f (a) and f (b), there is a number c between a and bsuch that f (c) = u.

Graphically speaking, this theorem just says that acontinuous curve cannot get from one level to anotherwithout crossing every level in between.

For example, consider the function f defined byf (x) = x2 − 2. f is continuous and f (−2) > 0 > f (−1);therefore, by the IVT, there is a number c between −2and −1 such that f (c) = 0. What is c?

This theorem depends on the continuity of the realnumber system. It would not be true in the rationalnumber system, for example.


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Exercises

1 Show that the equation x4 − x3 = 5 has a solutionbetween 1 and 2.

2 Show that the equation cos x = x has a solution between0 and 1.


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Exercises

1 Show that the equation x4 − x3 = 5 has a solutionbetween 1 and 2.Solution: The function defined by f (x) = x4 − x3, being apolynomial function, is continuous on [1, 2]. f (1) = 0 andf (2) = 8; therefore, by the IVT, there must be a number cbetween 1 and 2 such that f (c) = 5.

2 Show that the equation cos x = x has a solution between0 and 1.


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Exercises

1 Show that the equation x4 − x3 = 5 has a solutionbetween 1 and 2. Solution: The function defined byf (x) = x4 − x3, being a polynomial function, is continuouson [1, 2]. f (1) = 0 and f (2) = 8; therefore, by the IVT,there must be a number c between 1 and 2 such thatf (c) = 5.

2 Show that the equation cos x = x has a solution between0 and 1. Solution: Consider the function f defined byf (x) = cos x − x . Since f is a difference of continuousfunctions, f is continuous. Solving cos x = x is equivalentto finding a root of f . f (0) = 1− 0 = 1 > 0; since 1 < π

2 ,cos(1) < 1, and hence f (1) = cos(1)− 1 < 0. Therefore,by the Intermediate Value Theorem, there is a number cbetween 0 and 1 such that f (c) = 0.


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The End(of Chapter 1)


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Suggested Study Protocol & Tips

Finish all homework. (This should be done already!)

Reread the chapter from start to finish, or at least skim it.

Reread this presentation and, if necessary, the introductorybackground presentation, to help you mentally organizethe material. All class presentations are on my Web site.

Return to any homework problems or in-class examples youhad difficulty with and make sure you can do them now.

If you have any questions, come to office hours (or makean appointment, if necessary) and ask them!

Not right before the test!

Review the assignments as a whole to remind yourself ofthe types of problems you need to be able to do.

Take the sample test as if it were a real test.


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Suggested Study Protocol & Tips (Continued)

If your studying generates any further questions, come tooffice hours (or make an appointment, if necessary) andask them!

Not right before the test!

In all you do, make sure you understand what you aredoing and why you are doing it.

If you don’t know what it means to understand what youare doing and why you are doing it, come to office hours....

Get a good night’s sleep before the test so you can thinkclearly!


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In particular, make sure you:

Know and understand every definition. (To test yourunderstanding, explain what the definition means to aclassmate or friend.) You should be able to state everydefinition from memory, fill in a missing part of anydefinition, and interpret every definition in order to applyit or answer conceptual questions about it.Know and understand every theorem. (To test yourunderstanding, explain what the theorem means to aclassmate or friend.) You should be able to state everytheorem from memory, fill in a missing part of anytheorem, and interpret every theorem in order to apply itor answer conceptual questions about it.Understand every concept both verbally and graphically.Can do all the assigned computations and other exercises!Prepare, prepare, prepare! You cannot be overprepared;you can easily be underprepared!


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Good Luck!

But ... if you study well,you won’t need it,

and if you don’t study well,it won’t help you!

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