Section 2.2–3The Concept of Limit

Limit Laws

Math 1a

February 4, 2008

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Outline

The Concept of LimitHeuristicsErrors and tolerancesPathologies

Limit LawsEasy lawsThe direct substitution propertyLimits by algebraTwo more limit theorems

Zeno’s Paradox

That which is inlocomotion mustarrive at thehalf-way stagebefore it arrives atthe goal.

(Aristotle Physics VI:9,239b10)

Heuristic Definition of a Limit

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 toL as we like) by taking x to be sufficiently close to a (on either sideof a) but not equal to a.

The error-tolerance game

This tolerance is too bigStill too bigThis looks goodSo does this

a

L

The error-tolerance game

This tolerance is too bigStill too bigThis looks goodSo does this

a

L

The error-tolerance game

This tolerance is too bigStill too bigThis looks goodSo does this

a

L

The error-tolerance game

This tolerance is too big

Still too bigThis looks goodSo does this

a

L

The error-tolerance game

This tolerance is too bigStill too bigThis looks goodSo does this

a

L

The error-tolerance game

This tolerance is too big

Still too big

This looks goodSo does this

a

L

The error-tolerance game

This tolerance is too bigStill too bigThis looks goodSo does this

a

L

The error-tolerance game

This tolerance is too bigStill too big

This looks good

So does this

a

L

The error-tolerance game

This tolerance is too bigStill too bigThis looks good

So does this

a

L

Examples

Example

Find limx→0

x2 if it exists.

Example

Find limx→0

|x |x

if it exists.

Example

Find limx→0+

1

xif it exists.

Example

Find limx→0

sin(π

x

)if it exists.

Examples

Example

Find limx→0

x2 if it exists.

Example

Find limx→0

|x |x

if it exists.

Example

Find limx→0+

1

xif it exists.

Example

Find limx→0

sin(π

x

)if it exists.

Examples

Example

Find limx→0

x2 if it exists.

Example

Find limx→0

|x |x

if it exists.

Example

Find limx→0+

1

xif it exists.

Example

Find limx→0

sin(π

x

)if it exists.

Examples

Example

Find limx→0

x2 if it exists.

Example

Find limx→0

|x |x

if it exists.

Example

Find limx→0+

1

xif it exists.

Example

Find limx→0

sin(π

x

)if it exists.

What could go wrong?

How could a function fail to have a limit? Some possibilities:

I left- and right- hand limits exist but are not equal

I The function is unbounded near a

I Oscillation with increasingly high frequency near a

Precise Definition of a Limit

Let f be a function defined on an some open interval that containsthe number a, except possibly at a itself. Then we say that thelimit of f (x) as x approaches a is L, and we write

limx→a

f (x) = L,

if for every ε > 0 there is a corresponding δ > 0 such that

if 0 < |x − a| < δ, then |f (x)− L| < ε.

The error-tolerance game = ε, δ

L + ε

L− ε

a− δ a + δ

This δ is too big

a− δa + δ

This δ looks good

a− δa + δ

So does this δ

a

L

The error-tolerance game = ε, δ

L + ε

L− ε

a− δ a + δ

This δ is too big

a− δa + δ

This δ looks good

a− δa + δ

So does this δ

a

L

The error-tolerance game = ε, δ

L + ε

L− ε

a− δ a + δ

This δ is too big

a− δa + δ

This δ looks good

a− δa + δ

So does this δ

a

L

The error-tolerance game = ε, δ

L + ε

L− ε

a− δ a + δ

This δ is too big

a− δa + δ

This δ looks good

a− δa + δ

So does this δ

a

L

The error-tolerance game = ε, δ

L + ε

L− ε

a− δ a + δ

This δ is too big

a− δa + δ

This δ looks good

a− δa + δ

So does this δ

a

L

The error-tolerance game = ε, δ

L + ε

L− ε

a− δ a + δ

This δ is too big

a− δa + δ

This δ looks good

a− δa + δ

So does this δ

a

L

The error-tolerance game = ε, δ

L + ε

L− ε

a− δ a + δ

This δ is too big

a− δa + δ

This δ looks good

a− δa + δ

So does this δ

a

L

Meet the Mathematician: Augustin Louis Cauchy

I French, 1789–1857

I Royalist and Catholic

I made contributions ingeometry, calculus,complex analysis,number theory

I created the definition oflimit we use today butdidn’t understand it

Outline

The Concept of LimitHeuristicsErrors and tolerancesPathologies

Limit LawsEasy lawsThe direct substitution propertyLimits by algebraTwo more limit theorems

Limit Laws

Suppose that c is a constant and the limits

limx→a

f (x) and limx→a

g(x)

exist. Then

1. limx→a

[f (x) + g(x)] = limx→a

f (x) + limx→a

g(x)

2. limx→a

[f (x)− g(x)] = limx→a

f (x)− limx→a

g(x)

3. limx→a

[cf (x)] = c limx→a

f (x)

4. limx→a

[f (x)g(x)] = limx→a

f (x) · limx→a

g(x)

Limit Laws, continued

5. limx→a

f (x)

g(x)=

limx→a

f (x)

limx→a

g(x), if lim

x→ag(x) 6= 0.

6. limx→a

[f (x)]n =[

limx→a

f (x)]n

(follows from 3 repeatedly)

7. limx→a

c = c

8. limx→a

x = a

9. limx→a

xn = an

(follows from 6 and 8)

10. limx→a

n√

x = n√

a

11. limx→a

n√

f (x) = n

√limx→a

f (x) (If n is even, we must additionally

assume that limx→a

f (x) > 0)

Limit Laws, continued

5. limx→a

f (x)

g(x)=

limx→a

f (x)

limx→a

g(x), if lim

x→ag(x) 6= 0.

6. limx→a

[f (x)]n =[

limx→a

f (x)]n

(follows from 3 repeatedly)

7. limx→a

c = c

8. limx→a

x = a

9. limx→a

xn = an

(follows from 6 and 8)

10. limx→a

n√

x = n√

a

11. limx→a

n√

f (x) = n

√limx→a

f (x) (If n is even, we must additionally

assume that limx→a

f (x) > 0)

Limit Laws, continued

5. limx→a

f (x)

g(x)=

limx→a

f (x)

limx→a

g(x), if lim

x→ag(x) 6= 0.

6. limx→a

[f (x)]n =[

limx→a

f (x)]n

(follows from 3 repeatedly)

7. limx→a

c = c

8. limx→a

x = a

9. limx→a

xn = an

(follows from 6 and 8)

10. limx→a

n√

x = n√

a

11. limx→a

n√

f (x) = n

√limx→a

f (x) (If n is even, we must additionally

assume that limx→a

f (x) > 0)

Limit Laws, continued

5. limx→a

f (x)

g(x)=

limx→a

f (x)

limx→a

g(x), if lim

x→ag(x) 6= 0.

6. limx→a

[f (x)]n =[

limx→a

f (x)]n

(follows from 3 repeatedly)

7. limx→a

c = c

8. limx→a

x = a

9. limx→a

xn = an

(follows from 6 and 8)

10. limx→a

n√

x = n√

a

11. limx→a

n√

f (x) = n

√limx→a

f (x) (If n is even, we must additionally

assume that limx→a

f (x) > 0)

Limit Laws, continued

5. limx→a

f (x)

g(x)=

limx→a

f (x)

limx→a

g(x), if lim

x→ag(x) 6= 0.

6. limx→a

[f (x)]n =[

limx→a

f (x)]n

(follows from 3 repeatedly)

7. limx→a

c = c

8. limx→a

x = a

9. limx→a

xn = an

(follows from 6 and 8)

10. limx→a

n√

x = n√

a

11. limx→a

n√

f (x) = n

√limx→a

f (x) (If n is even, we must additionally

assume that limx→a

f (x) > 0)

Limit Laws, continued

5. limx→a

f (x)

g(x)=

limx→a

f (x)

limx→a

g(x), if lim

x→ag(x) 6= 0.

6. limx→a

[f (x)]n =[

limx→a

f (x)]n

(follows from 3 repeatedly)

7. limx→a

c = c

8. limx→a

x = a

9. limx→a

xn = an (follows from 6 and 8)

10. limx→a

n√

x = n√

a

11. limx→a

n√

f (x) = n

√limx→a

f (x) (If n is even, we must additionally

assume that limx→a

f (x) > 0)

Limit Laws, continued

5. limx→a

f (x)

g(x)=

limx→a

f (x)

limx→a

g(x), if lim

x→ag(x) 6= 0.

6. limx→a

[f (x)]n =[

limx→a

f (x)]n

(follows from 3 repeatedly)

7. limx→a

c = c

8. limx→a

x = a

9. limx→a

xn = an (follows from 6 and 8)

10. limx→a

n√

x = n√

a

11. limx→a

n√

f (x) = n

√limx→a

f (x) (If n is even, we must additionally

assume that limx→a

f (x) > 0)

Direct Substitution Property

Theorem (The Direct Substitution Property)

If f is a polynomial or a rational function and a is in the domain off , then

limx→a

f (x) = f (a)

Limits do not see the point! (in a good way)

TheoremIf f (x) = g(x) when x 6= a, and lim

x→ag(x) = L, then lim

x→af (x) = L.

Example

Find limx→−1

x2 + 2x + 1

x + 1, if it exists.

Solution

Sincex2 + 2x + 1

x + 1= x + 1 whenever x 6= −1, and since

limx→−1

x + 1 = 0, we have limx→−1

x2 + 2x + 1

x + 1= 0.

Limits do not see the point! (in a good way)

TheoremIf f (x) = g(x) when x 6= a, and lim

x→ag(x) = L, then lim

x→af (x) = L.

Example

Find limx→−1

x2 + 2x + 1

x + 1, if it exists.

Solution

Sincex2 + 2x + 1

x + 1= x + 1 whenever x 6= −1, and since

limx→−1

x + 1 = 0, we have limx→−1

x2 + 2x + 1

x + 1= 0.

Limits do not see the point! (in a good way)

TheoremIf f (x) = g(x) when x 6= a, and lim

x→ag(x) = L, then lim

x→af (x) = L.

Example

Find limx→−1

x2 + 2x + 1

x + 1, if it exists.

Solution

Sincex2 + 2x + 1

x + 1= x + 1 whenever x 6= −1, and since

limx→−1

x + 1 = 0, we have limx→−1

x2 + 2x + 1

x + 1= 0.

Finding limits by algebraic manipulations

Example

Find limx→4

√x − 2

x − 4.

SolutionWrite the denominator as x − 4 =

√x

2 − 4 = (√

x − 2)(√

x + 2).So

limx→2

√x − 2

x − 4= lim

x→2

√x − 2

(√

x − 2)(√

x + 2)

= limx→2

1√x + 2

=1

4

Example

Try limx→a

3√

x − 3√

a

x − a.

Finding limits by algebraic manipulations

Example

Find limx→4

√x − 2

x − 4.

SolutionWrite the denominator as x − 4 =

√x

2 − 4 = (√

x − 2)(√

x + 2).

So

limx→2

√x − 2

x − 4= lim

x→2

√x − 2

(√

x − 2)(√

x + 2)

= limx→2

1√x + 2

=1

4

Example

Try limx→a

3√

x − 3√

a

x − a.

Finding limits by algebraic manipulations

Example

Find limx→4

√x − 2

x − 4.

SolutionWrite the denominator as x − 4 =

√x

2 − 4 = (√

x − 2)(√

x + 2).So

limx→2

√x − 2

x − 4= lim

x→2

√x − 2

(√

x − 2)(√

x + 2)

= limx→2

1√x + 2

=1

4

Example

Try limx→a

3√

x − 3√

a

x − a.

Finding limits by algebraic manipulations

Example

Find limx→4

√x − 2

x − 4.

SolutionWrite the denominator as x − 4 =

√x

2 − 4 = (√

x − 2)(√

x + 2).So

limx→2

√x − 2

x − 4= lim

x→2

√x − 2

(√

x − 2)(√

x + 2)

= limx→2

1√x + 2

=1

4

Example

Try limx→a

3√

x − 3√

a

x − a.

Two More Important Limit Theorems

TheoremIf f (x) ≤ g(x) when x is near a (except possibly at a), then

limx→a

f (x) ≤ limx→a

g(x)

(as usual, provided these limits exist).

Theorem (The Squeeze/Sandwich/Pinching Theorem)

If f (x) ≤ g(x) ≤ h(x) when x is near a (as usual, except possiblyat a), and

limx→a

f (x) = limx→a

h(x) = L,

thenlimx→a

g(x) = L.

We can use the Squeeze Theorem to make complicated limitssimple.

Example

Show that limx→0

x2 sin

(1

x

)= 0.

SolutionWe have for all x,

−x2 ≤ x2 sin

(1

x

)≤ x2

The left and right sides go to zero as x → 0.

We can use the Squeeze Theorem to make complicated limitssimple.

Example

Show that limx→0

x2 sin

(1

x

)= 0.

SolutionWe have for all x,

−x2 ≤ x2 sin

(1

x

)≤ x2

The left and right sides go to zero as x → 0.

We can use the Squeeze Theorem to make complicated limitssimple.

Example

Show that limx→0

x2 sin

(1

x

)= 0.

SolutionWe have for all x,

−x2 ≤ x2 sin

(1

x

)≤ x2

The left and right sides go to zero as x → 0.