day 5 of the intuitive online calculus course: the squeeze theorem

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