Section 2.2–3 The Concept of Limit Limit Laws Math 1a February 4, 2008 Announcements I Syllabus available on course website I Homework for Wednesday 2/6: I Practice 2.2: 1, 3, 5, 7, 13, 15; 2.3: 1, 3, 7, 13, 15, 17 I Turn-in 2:2: 2, 4, 6, 8; 2.3: 2, 20, 38 I Homework for Monday 2/11: 2.2.28, 2.3.30, 2.4.34 I ALEKS due Wednesday 2/20
The concept of limit formalizes the notion of closeness of the function values to a certain value "near" a certain point. Limits behave well with respect to arithmetic--usually. Division by zero is always a problem, and we can't make conclusions about nonexistent limits!
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Section 2.2–3The Concept of Limit
Limit Laws
Math 1a
February 4, 2008
Announcements
I Syllabus available on course websiteI Homework for Wednesday 2/6:
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.