day 5 of the intuitive online calculus course: the squeeze theorem
DESCRIPTION
In this presentation we learn about the Squeeze Theorem. We first try to get the intuition behind it, why it must be true. Then we apply it to solve the Fundamental Trigonometric Limit. This limit is very important for solving other trigonometric limits. To solve this limit we use a little bit of geometry and then apply the Squeeze Theorem.TRANSCRIPT
The Squeeze Theorem
The Squeeze Theorem
Let’s say f and g are two functions such that one is always belowthe other. Let’s say that g is always above:
The Squeeze Theorem
Let’s say f and g are two functions such that one is always belowthe other. Let’s say that g is always above:
f (x) ≤ g(x)
The Squeeze Theorem
Let’s say f and g are two functions such that one is always belowthe other. Let’s say that g is always above:
f (x) ≤ g(x)
Also, let’s say that their limits are equal in some point:
The Squeeze Theorem
Let’s say f and g are two functions such that one is always belowthe other. Let’s say that g is always above:
f (x) ≤ g(x)
Also, let’s say that their limits are equal in some point:
limx→a
f (x) = limx→a
g(x) = L
The Squeeze Theorem
The Squeeze Theorem
Now, let’s suppose we squeeze a third function between them:
The Squeeze Theorem
Now, let’s suppose we squeeze a third function between them:
The Squeeze Theorem
Now, let’s suppose we squeeze a third function between them:
Let’s call this third function h:
The Squeeze Theorem
Now, let’s suppose we squeeze a third function between them:
Let’s call this third function h:
f (x) ≤ h(x) ≤ g(x)
The Squeeze Theorem
Now, let’s suppose we squeeze a third function between them:
The Squeeze Theorem
Now, let’s suppose we squeeze a third function between them:
The Squeze Theorem says that:
The Squeeze Theorem
Now, let’s suppose we squeeze a third function between them:
The Squeze Theorem says that:
limx→a
h(x) = L
The Fundamental Trigonometric Limit
The Fundamental Trigonometric Limit
We’re going to prove the following limit using the SqueezeTheorem
The Fundamental Trigonometric Limit
We’re going to prove the following limit using the SqueezeTheorem
limx→0
sin x
x= 1
The Fundamental Trigonometric Limit
The Fundamental Trigonometric Limit
The Fundamental Trigonometric Limit
The Fundamental Trigonometric Limit
∆MOA < Sector MOA < ∆COA
The Fundamental Trigonometric Limit
sin x =
The Fundamental Trigonometric Limit
sin x =BM
1
The Fundamental Trigonometric Limit
sin x =BM
1= BM
The Fundamental Trigonometric Limit
sin x =BM
1= BM
∆MOA =
The Fundamental Trigonometric Limit
sin x =BM
1= BM
∆MOA =1.BM
2
The Fundamental Trigonometric Limit
sin x =BM
1= BM
∆MOA =1.BM
2=
sin x
2
The Fundamental Trigonometric Limit
∆MOA < Sector MOA < ∆COA
The Fundamental Trigonometric Limit
����:
sin x2
∆MOA < Sector MOA < ∆COA
The Fundamental Trigonometric Limit
sin x
2< Sector MOA < ∆COA
The Fundamental Trigonometric Limit
The Fundamental Trigonometric Limit
tan x =
The Fundamental Trigonometric Limit
tan x =AC
1
The Fundamental Trigonometric Limit
tan x =AC
1= AC
The Fundamental Trigonometric Limit
tan x =AC
1= AC
∆COA =
The Fundamental Trigonometric Limit
tan x =AC
1= AC
∆COA =1.AC
2=
The Fundamental Trigonometric Limit
tan x =AC
1= AC
∆COA =1.AC
2=
tan x
2
The Fundamental Trigonometric Limit
tan x =AC
1= AC
sin x
2< Sector MOA < ∆COA
The Fundamental Trigonometric Limit
tan x =AC
1= AC
sin x
2< Sector MOA <���
�:tan x2
∆COA
The Fundamental Trigonometric Limit
tan x =AC
1= AC
sin x
2< Sector MOA <
tan x
2
The Fundamental Trigonometric Limit
The Fundamental Trigonometric Limit
The area of a circular sector is:
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
A(2π) = πr2
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
A(2π) = πr2
If we multiply both sides of this equation by θ2π :
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
A(2π) = πr2
If we multiply both sides of this equation by θ2π :
θ
2πA(2π)
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
A(2π) = πr2
If we multiply both sides of this equation by θ2π :
θ
2πA(2π) = A(
θ2π
2π)
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
A(2π) = πr2
If we multiply both sides of this equation by θ2π :
θ
2πA(2π) = A(
θ��2π
��2π)
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
A(2π) = πr2
If we multiply both sides of this equation by θ2π :
θ
2πA(2π) = A(
θ��2π
��2π) = A(θ) =
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
A(2π) = πr2
If we multiply both sides of this equation by θ2π :
θ
2πA(2π) = A(
θ2π
2π) = A(θ) =
θπr2
2π
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
A(2π) = πr2
If we multiply both sides of this equation by θ2π :
θ
2πA(2π) = A(
θ2π
2π) = A(θ) =
θ�πr2
2�π
The Fundamental Trigonometric Limit
The area of a circular sector is:
A =θr2
2
A simple proof is the following:
A(2π) = πr2
If we multiply both sides of this equation by θ2π :
θ
2πA(2π) = A(
θ2π
2π) = A(θ) =
θ�πr2
2�π=θr2
2
The Fundamental Trigonometric Limit
The Fundamental Trigonometric Limit
The Fundamental Trigonometric Limit
Now, we have a circle of radius 1 and angle x , so, our area is:
The Fundamental Trigonometric Limit
Now, we have a circle of radius 1 and angle x , so, our area is:
Sector MOA =
The Fundamental Trigonometric Limit
Now, we have a circle of radius 1 and angle x , so, our area is:
Sector MOA =x .12
2=
The Fundamental Trigonometric Limit
Now, we have a circle of radius 1 and angle x , so, our area is:
Sector MOA =x .12
2=
x
2
The Fundamental Trigonometric Limit
Now, we have a circle of radius 1 and angle x , so, our area is:
sin x
2< Sector MOA <
tan x
2
The Fundamental Trigonometric Limit
Now, we have a circle of radius 1 and angle x , so, our area is:
sin x
2<���
����:
x2
Sector MOA <tan x
2
The Fundamental Trigonometric Limit
Now, we have a circle of radius 1 and angle x , so, our area is:
sin x
2<
x
2<
tan x
2
The Fundamental Trigonometric Limit
Now, we have a circle of radius 1 and angle x , so, our area is:
sin x
2<
x
2<
tan x
2
sin x < x < tan x
The Fundamental Trigonometric Limit
The Fundamental Trigonometric Limit
sin x < x < tan x
The Fundamental Trigonometric Limit
sin x < x < tan x
1 <x
sin x<
1
cos x
The Fundamental Trigonometric Limit
sin x < x < tan x
1 <x
sin x<
1
cos x
1 >sin x
x> cos x
The Fundamental Trigonometric Limit
sin x < x < tan x
1 <x
sin x<
1
cos x
1 >sin x
x>���:
1cos x
The Fundamental Trigonometric Limit
sin x < x < tan x
1 <x
sin x<
1
cos x
1 >sin x
x>���:
1cos x
Conclusion:
The Fundamental Trigonometric Limit
sin x < x < tan x
1 <x
sin x<
1
cos x
1 >sin x
x>���:
1cos x
Conclusion:
limx→0
sin x
x= 1
